Find the integral $int _{1}^{e} (x ln x)^2 dx$.
$begingroup$
Find : $$int _{1}^{e} (x ln x)^2 ;dx.$$
My answer:
I have tried integration by parts with $u = x^2$ and $dv = (ln x)^2$ but I end up having the same integration another time!
I reversed the role of $u$ & $v$, but it also did not work?
Do you have any suggestions ?
calculus integration analysis
edited Nov 27 '18 at 10:14
Anurag A
26k12249
asked Nov 27 '18 at 9:42
hopefullyhopefully
271113
$endgroup$
add a comment |
$begingroup$
Find : $$int _{1}^{e} (x ln x)^2 ;dx.$$
My answer:
I have tried integration by parts with $u = x^2$ and $dv = (ln x)^2$ but I end up having the same integration another time!
I reversed the role of $u$ & $v$, but it also did not work?
Do you have any suggestions ?
calculus integration analysis
edited Nov 27 '18 at 10:14
Anurag A
26k12249
asked Nov 27 '18 at 9:42
hopefullyhopefully
271113
$endgroup$
add a comment |
$begingroup$
Find : $$int _{1}^{e} (x ln x)^2 ;dx.$$
My answer:
I have tried integration by parts with $u = x^2$ and $dv = (ln x)^2$ but I end up having the same integration another time!
I reversed the role of $u$ & $v$, but it also did not work?
Do you have any suggestions ?
calculus integration analysis
edited Nov 27 '18 at 10:14
Anurag A
26k12249
asked Nov 27 '18 at 9:42
hopefullyhopefully
271113
$endgroup$
Find : $$int _{1}^{e} (x ln x)^2 ;dx.$$
My answer:
I have tried integration by parts with $u = x^2$ and $dv = (ln x)^2$ but I end up having the same integration another time!
I reversed the role of $u$ & $v$, but it also did not work?
Do you have any suggestions ?
calculus integration analysis
calculus integration analysis
edited Nov 27 '18 at 10:14
Anurag A
26k12249
asked Nov 27 '18 at 9:42
hopefullyhopefully
271113
edited Nov 27 '18 at 10:14
Anurag A
26k12249
asked Nov 27 '18 at 9:42
hopefullyhopefully
271113
edited Nov 27 '18 at 10:14
Anurag A
26k12249
edited Nov 27 '18 at 10:14
Anurag A
26k12249
edited Nov 27 '18 at 10:14
Anurag A
26k12249
26k12249
asked Nov 27 '18 at 9:42
hopefullyhopefully
271113
asked Nov 27 '18 at 9:42
hopefullyhopefully
271113
asked Nov 27 '18 at 9:42
hopefullyhopefully
271113
271113
add a comment |
add a comment |
4 Answers
4
active
oldest
votes
$begingroup$
Hint:
Let $ln x=yimplies x=e^y,dx=e^y dy$
$$int_1^e(xln x)^2=int_0^1e^{3y}y^2 dy$$
Now $dfrac{d(e^{my}y^n)}{dy}=me^{my}y^n+e^{my}ny^{n-1}$
If $displaystyle I(n)=int e^{my}y^n dy,$
$$mI(n)+nI(n-1)=e^{my}y^n+K$$
Here $m=3,n=2$
answered Nov 27 '18 at 9:49
lab bhattacharjeelab bhattacharjee
225k15156274
$endgroup$
add a comment |
$begingroup$
Hint: make thus substitution $y =ln, x$ and then integrate by parts.
answered Nov 27 '18 at 9:48
Kavi Rama MurthyKavi Rama Murthy
56k42158
$endgroup$
add a comment |
$begingroup$
$newcommand{bbx}[1]{,bbox[15px,border:1px groove navy]{displaystyle{#1}},}
newcommand{braces}[1]{leftlbrace,{#1},rightrbrace}
newcommand{bracks}[1]{leftlbrack,{#1},rightrbrack}
newcommand{dd}{mathrm{d}}
newcommand{ds}[1]{displaystyle{#1}}
newcommand{expo}[1]{,mathrm{e}^{#1},}
newcommand{ic}{mathrm{i}}
newcommand{mc}[1]{mathcal{#1}}
newcommand{mrm}[1]{mathrm{#1}}
newcommand{pars}[1]{left(,{#1},right)}
newcommand{partiald}[3]{frac{partial^{#1} #2}{partial #3^{#1}}}
newcommand{root}[2]{,sqrt[#1]{,{#2},},}
newcommand{totald}[3]{frac{mathrm{d}^{#1} #2}{mathrm{d} #3^{#1}}}
newcommand{verts}[1]{leftvert,{#1},rightvert}$
begin{align}
int_{1}^{expo{}}x^{nu},dd x & =
left.{x^{nu +1} over nu + 1},rightvert_{ 1}^{ expo{}} =
{expo{nu +1} - 1 over nu + 1}
\[5mm] &
textsf{Derive twice respect of} nu:
\
int_{1}^{expo{}}x^{nu}ln^{2}pars{x},dd x & =
totald[2]{}{nu}pars{expo{nu +1} - 1 over nu + 1} =
{pars{nu^{2} + 1}expo{nu + 1} - 2 over pars{nu +1}^{3}}
\[5mm] &
textsf{Evaluate the limit} nu to 2:
\
int_{1}^{expo{}}bracks{xlnpars{x}}^{2},dd x & =
bbx{5expo{3} - 2 over 27} approx 3.6455
end{align}
answered Nov 27 '18 at 17:11
Felix MarinFelix Marin
67.5k7107141
$endgroup$
add a comment |
$begingroup$
Let $u=(ln x)^2$ and $dv=x^2 ,dx$. Then $du=2(ln x)frac{1}{x} , dx$ and $v=frac{x^3}{3}$. So
$$I=int_1^e(x ln x)^2 , dx=(ln x)^2 frac{x^3}{3}Big|_{1}^{e}-frac{2}{3}int_1^e x^2 ln x , dx.$$
Now we will solve the integral on the right side. Call the integral as $J$. For this $u=ln x$ and $dv=x^2 , dx$. So $du= frac{1}{x},dx$ and $v=frac{x^3}{3}$.Then
$$J=(ln x) frac{x^3}{3}Big|_{1}^{e}-frac{1}{3}int_1^e x^ 2, dx=(ln x) frac{x^3}{3}Big|_{1}^{e}-frac{x^3}{9}Big|_{1}^{e}=frac{e^3}{3}-left(frac{e^3-1}{9}right)=frac{2e^3+1}{9}.$$
So
$$I=frac{e^3}{3}-frac{2}{3}left(frac{2e^3+1}{9}right)=frac{5e^3-2}{27}$$
answered Nov 27 '18 at 9:54
Anurag AAnurag A
26k12249
$endgroup$
$begingroup$
My final answer is different from you ..... are you sure that yours is correct ?
$endgroup$
– hopefully
Nov 27 '18 at 10:10
$begingroup$
Mine is $frac {6 e^3 + 3}{27}$
$endgroup$
– hopefully
Nov 27 '18 at 10:11
$begingroup$
@hopefully I am quite sure my answer is correct.
$endgroup$
– Anurag A
Nov 27 '18 at 10:13
$begingroup$
the problem with me is your last line ..... all the previous is okay for me
$endgroup$
– hopefully
Nov 27 '18 at 10:18
1
$begingroup$
@hopefully you are mistaken $J$ has only $x^2 ln x$ (the log term is NOT squared) whereas your actual integral (which I am referring to as $I$) has $(x ln x)^2=x^2 (ln x)^2$.
$endgroup$
– Anurag A
Nov 27 '18 at 10:22
|
show 3 more comments
Your Answer
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4 Answers
4
active
oldest
votes
4 Answers
4
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
Hint:
Let $ln x=yimplies x=e^y,dx=e^y dy$
$$int_1^e(xln x)^2=int_0^1e^{3y}y^2 dy$$
Now $dfrac{d(e^{my}y^n)}{dy}=me^{my}y^n+e^{my}ny^{n-1}$
If $displaystyle I(n)=int e^{my}y^n dy,$
$$mI(n)+nI(n-1)=e^{my}y^n+K$$
Here $m=3,n=2$
answered Nov 27 '18 at 9:49
lab bhattacharjeelab bhattacharjee
225k15156274
$endgroup$
add a comment |
$begingroup$
Hint:
Let $ln x=yimplies x=e^y,dx=e^y dy$
$$int_1^e(xln x)^2=int_0^1e^{3y}y^2 dy$$
Now $dfrac{d(e^{my}y^n)}{dy}=me^{my}y^n+e^{my}ny^{n-1}$
If $displaystyle I(n)=int e^{my}y^n dy,$
$$mI(n)+nI(n-1)=e^{my}y^n+K$$
Here $m=3,n=2$
answered Nov 27 '18 at 9:49
lab bhattacharjeelab bhattacharjee
225k15156274
$endgroup$
add a comment |
$begingroup$
Hint:
Let $ln x=yimplies x=e^y,dx=e^y dy$
$$int_1^e(xln x)^2=int_0^1e^{3y}y^2 dy$$
Now $dfrac{d(e^{my}y^n)}{dy}=me^{my}y^n+e^{my}ny^{n-1}$
If $displaystyle I(n)=int e^{my}y^n dy,$
$$mI(n)+nI(n-1)=e^{my}y^n+K$$
Here $m=3,n=2$
answered Nov 27 '18 at 9:49
lab bhattacharjeelab bhattacharjee
225k15156274
$endgroup$
Hint:
Let $ln x=yimplies x=e^y,dx=e^y dy$
$$int_1^e(xln x)^2=int_0^1e^{3y}y^2 dy$$
Now $dfrac{d(e^{my}y^n)}{dy}=me^{my}y^n+e^{my}ny^{n-1}$
If $displaystyle I(n)=int e^{my}y^n dy,$
$$mI(n)+nI(n-1)=e^{my}y^n+K$$
Here $m=3,n=2$
answered Nov 27 '18 at 9:49
lab bhattacharjeelab bhattacharjee
225k15156274
answered Nov 27 '18 at 9:49
lab bhattacharjeelab bhattacharjee
225k15156274
answered Nov 27 '18 at 9:49
lab bhattacharjeelab bhattacharjee
225k15156274
answered Nov 27 '18 at 9:49
lab bhattacharjeelab bhattacharjee
225k15156274
225k15156274
add a comment |
add a comment |
$begingroup$
Hint: make thus substitution $y =ln, x$ and then integrate by parts.
answered Nov 27 '18 at 9:48
Kavi Rama MurthyKavi Rama Murthy
56k42158
$endgroup$
add a comment |
$begingroup$
Hint: make thus substitution $y =ln, x$ and then integrate by parts.
answered Nov 27 '18 at 9:48
Kavi Rama MurthyKavi Rama Murthy
56k42158
$endgroup$
add a comment |
$begingroup$
Hint: make thus substitution $y =ln, x$ and then integrate by parts.
answered Nov 27 '18 at 9:48
Kavi Rama MurthyKavi Rama Murthy
56k42158
$endgroup$
Hint: make thus substitution $y =ln, x$ and then integrate by parts.
answered Nov 27 '18 at 9:48
Kavi Rama MurthyKavi Rama Murthy
56k42158
answered Nov 27 '18 at 9:48
Kavi Rama MurthyKavi Rama Murthy
56k42158
answered Nov 27 '18 at 9:48
Kavi Rama MurthyKavi Rama Murthy
56k42158
answered Nov 27 '18 at 9:48
Kavi Rama MurthyKavi Rama Murthy
56k42158
56k42158
add a comment |
add a comment |
$begingroup$
$newcommand{bbx}[1]{,bbox[15px,border:1px groove navy]{displaystyle{#1}},}
newcommand{braces}[1]{leftlbrace,{#1},rightrbrace}
newcommand{bracks}[1]{leftlbrack,{#1},rightrbrack}
newcommand{dd}{mathrm{d}}
newcommand{ds}[1]{displaystyle{#1}}
newcommand{expo}[1]{,mathrm{e}^{#1},}
newcommand{ic}{mathrm{i}}
newcommand{mc}[1]{mathcal{#1}}
newcommand{mrm}[1]{mathrm{#1}}
newcommand{pars}[1]{left(,{#1},right)}
newcommand{partiald}[3]{frac{partial^{#1} #2}{partial #3^{#1}}}
newcommand{root}[2]{,sqrt[#1]{,{#2},},}
newcommand{totald}[3]{frac{mathrm{d}^{#1} #2}{mathrm{d} #3^{#1}}}
newcommand{verts}[1]{leftvert,{#1},rightvert}$
begin{align}
int_{1}^{expo{}}x^{nu},dd x & =
left.{x^{nu +1} over nu + 1},rightvert_{ 1}^{ expo{}} =
{expo{nu +1} - 1 over nu + 1}
\[5mm] &
textsf{Derive twice respect of} nu:
\
int_{1}^{expo{}}x^{nu}ln^{2}pars{x},dd x & =
totald[2]{}{nu}pars{expo{nu +1} - 1 over nu + 1} =
{pars{nu^{2} + 1}expo{nu + 1} - 2 over pars{nu +1}^{3}}
\[5mm] &
textsf{Evaluate the limit} nu to 2:
\
int_{1}^{expo{}}bracks{xlnpars{x}}^{2},dd x & =
bbx{5expo{3} - 2 over 27} approx 3.6455
end{align}
answered Nov 27 '18 at 17:11
Felix MarinFelix Marin
67.5k7107141
$endgroup$
add a comment |
$begingroup$
$newcommand{bbx}[1]{,bbox[15px,border:1px groove navy]{displaystyle{#1}},}
newcommand{braces}[1]{leftlbrace,{#1},rightrbrace}
newcommand{bracks}[1]{leftlbrack,{#1},rightrbrack}
newcommand{dd}{mathrm{d}}
newcommand{ds}[1]{displaystyle{#1}}
newcommand{expo}[1]{,mathrm{e}^{#1},}
newcommand{ic}{mathrm{i}}
newcommand{mc}[1]{mathcal{#1}}
newcommand{mrm}[1]{mathrm{#1}}
newcommand{pars}[1]{left(,{#1},right)}
newcommand{partiald}[3]{frac{partial^{#1} #2}{partial #3^{#1}}}
newcommand{root}[2]{,sqrt[#1]{,{#2},},}
newcommand{totald}[3]{frac{mathrm{d}^{#1} #2}{mathrm{d} #3^{#1}}}
newcommand{verts}[1]{leftvert,{#1},rightvert}$
begin{align}
int_{1}^{expo{}}x^{nu},dd x & =
left.{x^{nu +1} over nu + 1},rightvert_{ 1}^{ expo{}} =
{expo{nu +1} - 1 over nu + 1}
\[5mm] &
textsf{Derive twice respect of} nu:
\
int_{1}^{expo{}}x^{nu}ln^{2}pars{x},dd x & =
totald[2]{}{nu}pars{expo{nu +1} - 1 over nu + 1} =
{pars{nu^{2} + 1}expo{nu + 1} - 2 over pars{nu +1}^{3}}
\[5mm] &
textsf{Evaluate the limit} nu to 2:
\
int_{1}^{expo{}}bracks{xlnpars{x}}^{2},dd x & =
bbx{5expo{3} - 2 over 27} approx 3.6455
end{align}
answered Nov 27 '18 at 17:11
Felix MarinFelix Marin
67.5k7107141
$endgroup$
add a comment |
$begingroup$
$newcommand{bbx}[1]{,bbox[15px,border:1px groove navy]{displaystyle{#1}},}
newcommand{braces}[1]{leftlbrace,{#1},rightrbrace}
newcommand{bracks}[1]{leftlbrack,{#1},rightrbrack}
newcommand{dd}{mathrm{d}}
newcommand{ds}[1]{displaystyle{#1}}
newcommand{expo}[1]{,mathrm{e}^{#1},}
newcommand{ic}{mathrm{i}}
newcommand{mc}[1]{mathcal{#1}}
newcommand{mrm}[1]{mathrm{#1}}
newcommand{pars}[1]{left(,{#1},right)}
newcommand{partiald}[3]{frac{partial^{#1} #2}{partial #3^{#1}}}
newcommand{root}[2]{,sqrt[#1]{,{#2},},}
newcommand{totald}[3]{frac{mathrm{d}^{#1} #2}{mathrm{d} #3^{#1}}}
newcommand{verts}[1]{leftvert,{#1},rightvert}$
begin{align}
int_{1}^{expo{}}x^{nu},dd x & =
left.{x^{nu +1} over nu + 1},rightvert_{ 1}^{ expo{}} =
{expo{nu +1} - 1 over nu + 1}
\[5mm] &
textsf{Derive twice respect of} nu:
\
int_{1}^{expo{}}x^{nu}ln^{2}pars{x},dd x & =
totald[2]{}{nu}pars{expo{nu +1} - 1 over nu + 1} =
{pars{nu^{2} + 1}expo{nu + 1} - 2 over pars{nu +1}^{3}}
\[5mm] &
textsf{Evaluate the limit} nu to 2:
\
int_{1}^{expo{}}bracks{xlnpars{x}}^{2},dd x & =
bbx{5expo{3} - 2 over 27} approx 3.6455
end{align}
answered Nov 27 '18 at 17:11
Felix MarinFelix Marin
67.5k7107141
$endgroup$
$newcommand{bbx}[1]{,bbox[15px,border:1px groove navy]{displaystyle{#1}},}
newcommand{braces}[1]{leftlbrace,{#1},rightrbrace}
newcommand{bracks}[1]{leftlbrack,{#1},rightrbrack}
newcommand{dd}{mathrm{d}}
newcommand{ds}[1]{displaystyle{#1}}
newcommand{expo}[1]{,mathrm{e}^{#1},}
newcommand{ic}{mathrm{i}}
newcommand{mc}[1]{mathcal{#1}}
newcommand{mrm}[1]{mathrm{#1}}
newcommand{pars}[1]{left(,{#1},right)}
newcommand{partiald}[3]{frac{partial^{#1} #2}{partial #3^{#1}}}
newcommand{root}[2]{,sqrt[#1]{,{#2},},}
newcommand{totald}[3]{frac{mathrm{d}^{#1} #2}{mathrm{d} #3^{#1}}}
newcommand{verts}[1]{leftvert,{#1},rightvert}$
begin{align}
int_{1}^{expo{}}x^{nu},dd x & =
left.{x^{nu +1} over nu + 1},rightvert_{ 1}^{ expo{}} =
{expo{nu +1} - 1 over nu + 1}
\[5mm] &
textsf{Derive twice respect of} nu:
\
int_{1}^{expo{}}x^{nu}ln^{2}pars{x},dd x & =
totald[2]{}{nu}pars{expo{nu +1} - 1 over nu + 1} =
{pars{nu^{2} + 1}expo{nu + 1} - 2 over pars{nu +1}^{3}}
\[5mm] &
textsf{Evaluate the limit} nu to 2:
\
int_{1}^{expo{}}bracks{xlnpars{x}}^{2},dd x & =
bbx{5expo{3} - 2 over 27} approx 3.6455
end{align}
answered Nov 27 '18 at 17:11
Felix MarinFelix Marin
67.5k7107141
answered Nov 27 '18 at 17:11
Felix MarinFelix Marin
67.5k7107141
answered Nov 27 '18 at 17:11
Felix MarinFelix Marin
67.5k7107141
answered Nov 27 '18 at 17:11
Felix MarinFelix Marin
67.5k7107141
67.5k7107141
add a comment |
add a comment |
$begingroup$
Let $u=(ln x)^2$ and $dv=x^2 ,dx$. Then $du=2(ln x)frac{1}{x} , dx$ and $v=frac{x^3}{3}$. So
$$I=int_1^e(x ln x)^2 , dx=(ln x)^2 frac{x^3}{3}Big|_{1}^{e}-frac{2}{3}int_1^e x^2 ln x , dx.$$
Now we will solve the integral on the right side. Call the integral as $J$. For this $u=ln x$ and $dv=x^2 , dx$. So $du= frac{1}{x},dx$ and $v=frac{x^3}{3}$.Then
$$J=(ln x) frac{x^3}{3}Big|_{1}^{e}-frac{1}{3}int_1^e x^ 2, dx=(ln x) frac{x^3}{3}Big|_{1}^{e}-frac{x^3}{9}Big|_{1}^{e}=frac{e^3}{3}-left(frac{e^3-1}{9}right)=frac{2e^3+1}{9}.$$
So
$$I=frac{e^3}{3}-frac{2}{3}left(frac{2e^3+1}{9}right)=frac{5e^3-2}{27}$$
answered Nov 27 '18 at 9:54
Anurag AAnurag A
26k12249
$endgroup$
$begingroup$
My final answer is different from you ..... are you sure that yours is correct ?
$endgroup$
– hopefully
Nov 27 '18 at 10:10
$begingroup$
Mine is $frac {6 e^3 + 3}{27}$
$endgroup$
– hopefully
Nov 27 '18 at 10:11
$begingroup$
@hopefully I am quite sure my answer is correct.
$endgroup$
– Anurag A
Nov 27 '18 at 10:13
$begingroup$
the problem with me is your last line ..... all the previous is okay for me
$endgroup$
– hopefully
Nov 27 '18 at 10:18
1
$begingroup$
@hopefully you are mistaken $J$ has only $x^2 ln x$ (the log term is NOT squared) whereas your actual integral (which I am referring to as $I$) has $(x ln x)^2=x^2 (ln x)^2$.
$endgroup$
– Anurag A
Nov 27 '18 at 10:22
|
show 3 more comments
$begingroup$
Let $u=(ln x)^2$ and $dv=x^2 ,dx$. Then $du=2(ln x)frac{1}{x} , dx$ and $v=frac{x^3}{3}$. So
$$I=int_1^e(x ln x)^2 , dx=(ln x)^2 frac{x^3}{3}Big|_{1}^{e}-frac{2}{3}int_1^e x^2 ln x , dx.$$
Now we will solve the integral on the right side. Call the integral as $J$. For this $u=ln x$ and $dv=x^2 , dx$. So $du= frac{1}{x},dx$ and $v=frac{x^3}{3}$.Then
$$J=(ln x) frac{x^3}{3}Big|_{1}^{e}-frac{1}{3}int_1^e x^ 2, dx=(ln x) frac{x^3}{3}Big|_{1}^{e}-frac{x^3}{9}Big|_{1}^{e}=frac{e^3}{3}-left(frac{e^3-1}{9}right)=frac{2e^3+1}{9}.$$
So
$$I=frac{e^3}{3}-frac{2}{3}left(frac{2e^3+1}{9}right)=frac{5e^3-2}{27}$$
answered Nov 27 '18 at 9:54
Anurag AAnurag A
26k12249
$endgroup$
$begingroup$
My final answer is different from you ..... are you sure that yours is correct ?
$endgroup$
– hopefully
Nov 27 '18 at 10:10
$begingroup$
Mine is $frac {6 e^3 + 3}{27}$
$endgroup$
– hopefully
Nov 27 '18 at 10:11
$begingroup$
@hopefully I am quite sure my answer is correct.
$endgroup$
– Anurag A
Nov 27 '18 at 10:13
$begingroup$
the problem with me is your last line ..... all the previous is okay for me
$endgroup$
– hopefully
Nov 27 '18 at 10:18
1
$begingroup$
@hopefully you are mistaken $J$ has only $x^2 ln x$ (the log term is NOT squared) whereas your actual integral (which I am referring to as $I$) has $(x ln x)^2=x^2 (ln x)^2$.
$endgroup$
– Anurag A
Nov 27 '18 at 10:22
|
show 3 more comments
$begingroup$
Let $u=(ln x)^2$ and $dv=x^2 ,dx$. Then $du=2(ln x)frac{1}{x} , dx$ and $v=frac{x^3}{3}$. So
$$I=int_1^e(x ln x)^2 , dx=(ln x)^2 frac{x^3}{3}Big|_{1}^{e}-frac{2}{3}int_1^e x^2 ln x , dx.$$
Now we will solve the integral on the right side. Call the integral as $J$. For this $u=ln x$ and $dv=x^2 , dx$. So $du= frac{1}{x},dx$ and $v=frac{x^3}{3}$.Then
$$J=(ln x) frac{x^3}{3}Big|_{1}^{e}-frac{1}{3}int_1^e x^ 2, dx=(ln x) frac{x^3}{3}Big|_{1}^{e}-frac{x^3}{9}Big|_{1}^{e}=frac{e^3}{3}-left(frac{e^3-1}{9}right)=frac{2e^3+1}{9}.$$
So
$$I=frac{e^3}{3}-frac{2}{3}left(frac{2e^3+1}{9}right)=frac{5e^3-2}{27}$$
answered Nov 27 '18 at 9:54
Anurag AAnurag A
26k12249
$endgroup$
Let $u=(ln x)^2$ and $dv=x^2 ,dx$. Then $du=2(ln x)frac{1}{x} , dx$ and $v=frac{x^3}{3}$. So
$$I=int_1^e(x ln x)^2 , dx=(ln x)^2 frac{x^3}{3}Big|_{1}^{e}-frac{2}{3}int_1^e x^2 ln x , dx.$$
Now we will solve the integral on the right side. Call the integral as $J$. For this $u=ln x$ and $dv=x^2 , dx$. So $du= frac{1}{x},dx$ and $v=frac{x^3}{3}$.Then
$$J=(ln x) frac{x^3}{3}Big|_{1}^{e}-frac{1}{3}int_1^e x^ 2, dx=(ln x) frac{x^3}{3}Big|_{1}^{e}-frac{x^3}{9}Big|_{1}^{e}=frac{e^3}{3}-left(frac{e^3-1}{9}right)=frac{2e^3+1}{9}.$$
So
$$I=frac{e^3}{3}-frac{2}{3}left(frac{2e^3+1}{9}right)=frac{5e^3-2}{27}$$
answered Nov 27 '18 at 9:54
Anurag AAnurag A
26k12249
edited Nov 27 '18 at 10:00
answered Nov 27 '18 at 9:54
Anurag AAnurag A
26k12249
answered Nov 27 '18 at 9:54
Anurag AAnurag A
26k12249
answered Nov 27 '18 at 9:54
Anurag AAnurag A
26k12249
26k12249
$begingroup$
My final answer is different from you ..... are you sure that yours is correct ?
$endgroup$
– hopefully
Nov 27 '18 at 10:10
$begingroup$
Mine is $frac {6 e^3 + 3}{27}$
$endgroup$
– hopefully
Nov 27 '18 at 10:11
$begingroup$
@hopefully I am quite sure my answer is correct.
$endgroup$
– Anurag A
Nov 27 '18 at 10:13
$begingroup$
the problem with me is your last line ..... all the previous is okay for me
$endgroup$
– hopefully
Nov 27 '18 at 10:18
1
$begingroup$
@hopefully you are mistaken $J$ has only $x^2 ln x$ (the log term is NOT squared) whereas your actual integral (which I am referring to as $I$) has $(x ln x)^2=x^2 (ln x)^2$.
$endgroup$
– Anurag A
Nov 27 '18 at 10:22
|
show 3 more comments
$begingroup$
My final answer is different from you ..... are you sure that yours is correct ?
$endgroup$
– hopefully
Nov 27 '18 at 10:10
$begingroup$
Mine is $frac {6 e^3 + 3}{27}$
$endgroup$
– hopefully
Nov 27 '18 at 10:11
$begingroup$
@hopefully I am quite sure my answer is correct.
$endgroup$
– Anurag A
Nov 27 '18 at 10:13
$begingroup$
the problem with me is your last line ..... all the previous is okay for me
$endgroup$
– hopefully
Nov 27 '18 at 10:18
1
$begingroup$
@hopefully you are mistaken $J$ has only $x^2 ln x$ (the log term is NOT squared) whereas your actual integral (which I am referring to as $I$) has $(x ln x)^2=x^2 (ln x)^2$.
$endgroup$
– Anurag A
Nov 27 '18 at 10:22
$begingroup$
My final answer is different from you ..... are you sure that yours is correct ?
$endgroup$
– hopefully
Nov 27 '18 at 10:10
$begingroup$
My final answer is different from you ..... are you sure that yours is correct ?
$endgroup$
– hopefully
Nov 27 '18 at 10:10
$begingroup$
Mine is $frac {6 e^3 + 3}{27}$
$endgroup$
– hopefully
Nov 27 '18 at 10:11
$begingroup$
Mine is $frac {6 e^3 + 3}{27}$
$endgroup$
– hopefully
Nov 27 '18 at 10:11
$begingroup$
@hopefully I am quite sure my answer is correct.
$endgroup$
– Anurag A
Nov 27 '18 at 10:13
$begingroup$
@hopefully I am quite sure my answer is correct.
$endgroup$
– Anurag A
Nov 27 '18 at 10:13
$begingroup$
the problem with me is your last line ..... all the previous is okay for me
$endgroup$
– hopefully
Nov 27 '18 at 10:18
$begingroup$
the problem with me is your last line ..... all the previous is okay for me
$endgroup$
– hopefully
Nov 27 '18 at 10:18
1
1
$begingroup$
@hopefully you are mistaken $J$ has only $x^2 ln x$ (the log term is NOT squared) whereas your actual integral (which I am referring to as $I$) has $(x ln x)^2=x^2 (ln x)^2$.
$endgroup$
– Anurag A
Nov 27 '18 at 10:22
$begingroup$
@hopefully you are mistaken $J$ has only $x^2 ln x$ (the log term is NOT squared) whereas your actual integral (which I am referring to as $I$) has $(x ln x)^2=x^2 (ln x)^2$.
$endgroup$
– Anurag A
Nov 27 '18 at 10:22
|
show 3 more comments
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