Find int \ x^5/(1+x^2)^4 \ dx ?
1 Answer
int \ x^5/(1+x^2)^4 \ dx = - (3x^4+3x^2+1)/(6(1+x^2)^3) + C
Explanation:
We want to find:
I = int \ x^5/(1+x^2)^4 \ dx
Let us perform a substitution:
Let
u=1+x^2 => (du)/dx=2x; \ \ \ x^2=u-1
We can rewrite the integral and substitute as follows
I = int \ 1/2 (x^2)^2/(1+x^2)^4 \ (2x) \ dx
\ \ = 1/2 \ int \ (u-1)^2/(u)^4 \ du
\ \ = 1/2 \ int \ (u^2-2u+1)/(u^4) \ du
\ \ = 1/2 \ int \ 1/u^2-2/u^3+1/(u^4) \ du
Which we can now integrate to get:
I = 1/2 (-1/u +1/u^2 - 1 /(3u^3) ) + C
\ \ = -1/2 (1/u -1/u^2 + 1 /(3u^3) ) + C
\ \ = -1/2 ( (3u^2-3u+1)/(3u^3) ) + C
\ \ = -1/6 ( (3u^2-3u+1)/(u^3) ) + C
And by restoring the substitution we get:
I = -1/6 ( (3(1+x^2)^2-3(1+x^2)+1)/((1+x^2)^3) ) + C
\ \ = -1/6 ( (3x^4+6x^2+3-3-3x^2+1)/((1+x^2)^3) ) + C
\ \ = -1/6 ( (3x^4+3x^2+1)/((1+x^2)^3) ) + C
\ \ = -1/6 ( (3x^4+3x^2+1)/((1+x^2)^3) ) + C
