Question

Question #d53c9

Answer 1

`1/4 x^4 - 1/2x^2 + 1/2 ln |x^2+1| + C`

Explanation:

Using long division: `(x^5)/(x^2+1) = x^3-x + x/(x^2+1)`

So `int (x^5)/(x^2+1) dx = int (x^3-x) dx + int x/(x^2+1) dx`

For the last piece, let `u = x^2+1, du = 2x dx`

`int (x^5)/(x^2+1) dx = 1/4 x^4 - 1/2x^2 + 1/2 int (du)/u`

`int (x^5)/(x^2+1) dx = 1/4 x^4 - 1/2x^2 + 1/2 ln |x^2+1| + C`

Simplified:

`int (x^5)/(x^2+1) dx = 1/4 x^2(x^2-2)+ 1/2 ln |x^2+1| + C`

Answer 2

`intx^5/(x^2+1)dx = x^4/4 -x^2/2 + 1/2ln(x^2+1)+C`

Explanation:

Given:

`intx^5/(x^2+1)dx =`

Let `u = x^2", then "du = 2xdx`:

`1/2intu^2/(u+1)du=`

`1/2int(u^2+u-u)/(u+1)du=`

`1/2int(u^2+u)/(u+1)-(u)/(u+1)du=`

`1/2intu -(u)/(u+1)du=`

`1/2intu -(u+ 1 - 1)/(u+1)du=`

`1/2intu -(u+ 1)/(u+1) + 1/(u+1)du=`

`1/2intu -1 + 1/(u+1)du=`

`1/2(u^2/2 -u + ln|u+1|)+ C=`

Reverse the substitution:

`1/2(x^4/2 -x^2 + ln(x^2+1))+ C=`

`x^4/4 -x^2/2 + 1/2ln(x^2+1)+C`