Question

Question #234b7

Answer 1

I got: `ln(x^2+4)+c`

Explanation:

I would try setting:
`x^2+4=t`
so
`x^2=t-4`
`2xdx=dt`
substituting:
`intdt/t=ln|t|+c`
but `t=x^2+4`
and so the result is:
`ln(x^2+4)+c`

[`x^2+4` is always `>0`]