Question

Question #73ed7

Answer 1

`1/5x^5+8/3x^3+16x+C`

Explanation:

This looks like a u-substitution problem but you actually need to expand and integrate:

Start:
`int(x^2+4)^2dx`

Expand the integrand:
`=int(x^4+8x^2+16)dx`

Reverse the power rule:
`= 1/5x^5+8/3x^3+16x+C`

Answer 2

`1/5 x^5 + 8/3 x^3 + 16x + c`

Explanation:

My natural inteinct is using something called `u` substituion, but we can just expand...

`(x^2+4)^2 -= x^4 + 4x^2 + 4x^2 + 16`

`=> -= x^4 + 8x^2 + 16`

Now we have `int x^4 + 8x^2 + 16` `dx`

Now we use `int x^n dx = (x^(n+1))/(n+1) + c`

So our integral:

`color(red)(=> 1/5 x^5 + 8/3 x^3 + 16x + c`