Solve $intfrac{2x-3}{(x^2+x+1)^2}dx$
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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 at 13:34
user984325
15812
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up vote
2
down vote
favorite
$intfrac{2x-3}{(x^2+x+1)^2}dx$
$intfrac{2x-3}{(x^2+x+1)^2}dx=intfrac{2x+1}{(x^2+x+1)^2}dx-intfrac{4}{(x^2+x+1)^2}dx$
First integral is easily integrable but substituting $x^2+x+1=t$ but i cannot integrate the second integral.
integration
asked Nov 23 at 13:34
user984325
15812
add a comment |
up vote
2
down vote
favorite
up vote
2
down vote
favorite
$intfrac{2x-3}{(x^2+x+1)^2}dx$
$intfrac{2x-3}{(x^2+x+1)^2}dx=intfrac{2x+1}{(x^2+x+1)^2}dx-intfrac{4}{(x^2+x+1)^2}dx$
First integral is easily integrable but substituting $x^2+x+1=t$ but i cannot integrate the second integral.
integration
asked Nov 23 at 13:34
user984325
15812
$intfrac{2x-3}{(x^2+x+1)^2}dx$
$intfrac{2x-3}{(x^2+x+1)^2}dx=intfrac{2x+1}{(x^2+x+1)^2}dx-intfrac{4}{(x^2+x+1)^2}dx$
First integral is easily integrable but substituting $x^2+x+1=t$ but i cannot integrate the second integral.
integration
integration
asked Nov 23 at 13:34
user984325
15812
asked Nov 23 at 13:34
user984325
15812
asked Nov 23 at 13:34
user984325
15812
asked Nov 23 at 13:34
user984325
15812
asked Nov 23 at 13:34
user984325
15812
15812
add a comment |
add a comment |
2 Answers
2
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oldest
votes
up vote
5
down vote
accepted
Hint:
As $x^2+x+1=dfrac{(2x+1)^2+3}4,$ set $2x+1=sqrt3tan t$
answered Nov 23 at 13:36
lab bhattacharjee
222k15155273
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up vote
3
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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 at 13:48
lab bhattacharjee
222k15155273
add a comment |
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2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
5
down vote
accepted
Hint:
As $x^2+x+1=dfrac{(2x+1)^2+3}4,$ set $2x+1=sqrt3tan t$
answered Nov 23 at 13:36
lab bhattacharjee
222k15155273
add a comment |
up vote
5
down vote
accepted
Hint:
As $x^2+x+1=dfrac{(2x+1)^2+3}4,$ set $2x+1=sqrt3tan t$
answered Nov 23 at 13:36
lab bhattacharjee
222k15155273
add a comment |
up vote
5
down vote
accepted
up vote
5
down vote
accepted
Hint:
As $x^2+x+1=dfrac{(2x+1)^2+3}4,$ set $2x+1=sqrt3tan t$
answered Nov 23 at 13:36
lab bhattacharjee
222k15155273
Hint:
As $x^2+x+1=dfrac{(2x+1)^2+3}4,$ set $2x+1=sqrt3tan t$
answered Nov 23 at 13:36
lab bhattacharjee
222k15155273
answered Nov 23 at 13:36
lab bhattacharjee
222k15155273
answered Nov 23 at 13:36
lab bhattacharjee
222k15155273
answered Nov 23 at 13:36
lab bhattacharjee
222k15155273
222k15155273
add a comment |
add a comment |
up vote
3
down vote
$$dfrac{dleft(dfrac{ax^2+bx+c}{x^2+x+1}right)}{dx}=dfrac{(2ax+b)(x^2+x+1)-(ax^2+bx+c)(2x+1)}{(x^2+x+1)^2}$$
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 at 13:48
lab bhattacharjee
222k15155273
add a comment |
up vote
3
down vote
$$dfrac{dleft(dfrac{ax^2+bx+c}{x^2+x+1}right)}{dx}=dfrac{(2ax+b)(x^2+x+1)-(ax^2+bx+c)(2x+1)}{(x^2+x+1)^2}$$
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 at 13:48
lab bhattacharjee
222k15155273
add a comment |
up vote
3
down vote
up vote
3
down vote
$$dfrac{dleft(dfrac{ax^2+bx+c}{x^2+x+1}right)}{dx}=dfrac{(2ax+b)(x^2+x+1)-(ax^2+bx+c)(2x+1)}{(x^2+x+1)^2}$$
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 at 13:48
lab bhattacharjee
222k15155273
$$dfrac{dleft(dfrac{ax^2+bx+c}{x^2+x+1}right)}{dx}=dfrac{(2ax+b)(x^2+x+1)-(ax^2+bx+c)(2x+1)}{(x^2+x+1)^2}$$
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 at 13:48
lab bhattacharjee
222k15155273
answered Nov 23 at 13:48
lab bhattacharjee
222k15155273
answered Nov 23 at 13:48
lab bhattacharjee
222k15155273
answered Nov 23 at 13:48
lab bhattacharjee
222k15155273
222k15155273
add a comment |
add a comment |
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