Solve $intfrac{2x-3}{(x^2+x+1)^2}dx$
$intfrac{2x-3}{(x^2+x+1)^2}dx$
$intfrac{2x-3}{(x^2+x+1)^2}dx=intfrac{2x+1}{(x^2+x+1)^2}dx-intfrac{4}{(x^2+x+1)^2}dx$
First integral is easily integrable but substituting $x^2+x+1=t$ but i cannot integrate the second integral.
integration
asked Nov 23 '18 at 13:34
user984325user984325
246112
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$intfrac{2x-3}{(x^2+x+1)^2}dx$
$intfrac{2x-3}{(x^2+x+1)^2}dx=intfrac{2x+1}{(x^2+x+1)^2}dx-intfrac{4}{(x^2+x+1)^2}dx$
First integral is easily integrable but substituting $x^2+x+1=t$ but i cannot integrate the second integral.
integration
asked Nov 23 '18 at 13:34
user984325user984325
246112
add a comment |
$intfrac{2x-3}{(x^2+x+1)^2}dx$
$intfrac{2x-3}{(x^2+x+1)^2}dx=intfrac{2x+1}{(x^2+x+1)^2}dx-intfrac{4}{(x^2+x+1)^2}dx$
First integral is easily integrable but substituting $x^2+x+1=t$ but i cannot integrate the second integral.
integration
asked Nov 23 '18 at 13:34
user984325user984325
246112
$intfrac{2x-3}{(x^2+x+1)^2}dx$
$intfrac{2x-3}{(x^2+x+1)^2}dx=intfrac{2x+1}{(x^2+x+1)^2}dx-intfrac{4}{(x^2+x+1)^2}dx$
First integral is easily integrable but substituting $x^2+x+1=t$ but i cannot integrate the second integral.
integration
integration
asked Nov 23 '18 at 13:34
user984325user984325
246112
asked Nov 23 '18 at 13:34
user984325user984325
246112
asked Nov 23 '18 at 13:34
user984325user984325
246112
asked Nov 23 '18 at 13:34
user984325user984325
246112
asked Nov 23 '18 at 13:34
user984325user984325
246112
246112
add a comment |
add a comment |
2 Answers
2
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oldest
votes
Hint:
As $x^2+x+1=dfrac{(2x+1)^2+3}4,$ set $2x+1=sqrt3tan t$
answered Nov 23 '18 at 13:36
lab bhattacharjeelab bhattacharjee
224k15156274
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$$dfrac{dleft(dfrac{ax^2+bx+c}{x^2+x+1}right)}{dx}=dfrac{(2ax+b)(x^2+x+1)-(ax^2+bx+c)(2x+1)}{(x^2+x+1)^2}$$
The numerator $(2ax+b)(x^2+x+1)-(ax^2+bx+c)(2x+1)=x^2(a-b)+x(2a+2c)+b-c$
If the numerator $2x-3,$
$a-b=0iff a=b$
$b-c=-3iff c=b+3$
$2(a+c)=2iff1=a+c=b+b+3iff b=-1$
answered Nov 23 '18 at 13:48
lab bhattacharjeelab bhattacharjee
224k15156274
add a comment |
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2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
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active
oldest
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active
oldest
votes
Hint:
As $x^2+x+1=dfrac{(2x+1)^2+3}4,$ set $2x+1=sqrt3tan t$
answered Nov 23 '18 at 13:36
lab bhattacharjeelab bhattacharjee
224k15156274
add a comment |
Hint:
As $x^2+x+1=dfrac{(2x+1)^2+3}4,$ set $2x+1=sqrt3tan t$
answered Nov 23 '18 at 13:36
lab bhattacharjeelab bhattacharjee
224k15156274
add a comment |
Hint:
As $x^2+x+1=dfrac{(2x+1)^2+3}4,$ set $2x+1=sqrt3tan t$
answered Nov 23 '18 at 13:36
lab bhattacharjeelab bhattacharjee
224k15156274
Hint:
As $x^2+x+1=dfrac{(2x+1)^2+3}4,$ set $2x+1=sqrt3tan t$
answered Nov 23 '18 at 13:36
lab bhattacharjeelab bhattacharjee
224k15156274
answered Nov 23 '18 at 13:36
lab bhattacharjeelab bhattacharjee
224k15156274
answered Nov 23 '18 at 13:36
lab bhattacharjeelab bhattacharjee
224k15156274
answered Nov 23 '18 at 13:36
lab bhattacharjeelab bhattacharjee
224k15156274
224k15156274
add a comment |
add a comment |
$$dfrac{dleft(dfrac{ax^2+bx+c}{x^2+x+1}right)}{dx}=dfrac{(2ax+b)(x^2+x+1)-(ax^2+bx+c)(2x+1)}{(x^2+x+1)^2}$$
The numerator $(2ax+b)(x^2+x+1)-(ax^2+bx+c)(2x+1)=x^2(a-b)+x(2a+2c)+b-c$
If the numerator $2x-3,$
$a-b=0iff a=b$
$b-c=-3iff c=b+3$
$2(a+c)=2iff1=a+c=b+b+3iff b=-1$
answered Nov 23 '18 at 13:48
lab bhattacharjeelab bhattacharjee
224k15156274
add a comment |
$$dfrac{dleft(dfrac{ax^2+bx+c}{x^2+x+1}right)}{dx}=dfrac{(2ax+b)(x^2+x+1)-(ax^2+bx+c)(2x+1)}{(x^2+x+1)^2}$$
The numerator $(2ax+b)(x^2+x+1)-(ax^2+bx+c)(2x+1)=x^2(a-b)+x(2a+2c)+b-c$
If the numerator $2x-3,$
$a-b=0iff a=b$
$b-c=-3iff c=b+3$
$2(a+c)=2iff1=a+c=b+b+3iff b=-1$
answered Nov 23 '18 at 13:48
lab bhattacharjeelab bhattacharjee
224k15156274
add a comment |
$$dfrac{dleft(dfrac{ax^2+bx+c}{x^2+x+1}right)}{dx}=dfrac{(2ax+b)(x^2+x+1)-(ax^2+bx+c)(2x+1)}{(x^2+x+1)^2}$$
The numerator $(2ax+b)(x^2+x+1)-(ax^2+bx+c)(2x+1)=x^2(a-b)+x(2a+2c)+b-c$
If the numerator $2x-3,$
$a-b=0iff a=b$
$b-c=-3iff c=b+3$
$2(a+c)=2iff1=a+c=b+b+3iff b=-1$
answered Nov 23 '18 at 13:48
lab bhattacharjeelab bhattacharjee
224k15156274
$$dfrac{dleft(dfrac{ax^2+bx+c}{x^2+x+1}right)}{dx}=dfrac{(2ax+b)(x^2+x+1)-(ax^2+bx+c)(2x+1)}{(x^2+x+1)^2}$$
The numerator $(2ax+b)(x^2+x+1)-(ax^2+bx+c)(2x+1)=x^2(a-b)+x(2a+2c)+b-c$
If the numerator $2x-3,$
$a-b=0iff a=b$
$b-c=-3iff c=b+3$
$2(a+c)=2iff1=a+c=b+b+3iff b=-1$
answered Nov 23 '18 at 13:48
lab bhattacharjeelab bhattacharjee
224k15156274
answered Nov 23 '18 at 13:48
lab bhattacharjeelab bhattacharjee
224k15156274
answered Nov 23 '18 at 13:48
lab bhattacharjeelab bhattacharjee
224k15156274
answered Nov 23 '18 at 13:48
lab bhattacharjeelab bhattacharjee
224k15156274
224k15156274
add a comment |
add a comment |
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