Question

Question #37f63

Answer 1

`int x^3/(x^8+1)*dx=1/4arctan(x^4)+C`

Explanation:

`int x^3/(x^8+1)*dx`

=`1/4int (4x^3)/(x^8+1)*dx`

=`1/4arctan(x^4)+C`

Answer 2

`intx^3/(x^8+1)dx=1/4arctan(x^4)+"C"`

Explanation:

First make a substitution:

Let `color(blue)(u=x^4)->du=4x^3` `dx->color(red)(1/4du=x^3` `color(red)dx`

Given the integral

`intcolor(red)(x^3)/(x^8+1)` `color(red)dx`

We can rewrite the integral as

`color(red)(1/4)int 1/(color(blue)(u)^2+1)color(red)(du)`

Looking at only `int1/(u^2+1)du` we have a common integral whose antiderivative is `arctan(u)+"C"`

Thus, we now have

`color(red)(1/4)arctan(color(blue)(u))+"C"`

Finally, we substitute `x^4` back in for `color(blue)(u)` to get

`1/4arctan(x^4)+"C"`

`:.intx^3/(x^8+1)dx=1/4arctan(x^4)+"C"`