Question

How do you integrate `int xsqrt(2x^2+7)` using substitution?

Answer 1

I got: `intxsqrt(2x^2+7)dx=1/6(2x^2+7)sqrt(2x^2+7)+c`

Explanation:

Let us set:
`2x^2+7=t`
derive:
`4x=dt`
let us use this inside our integral (red):
`int1/4color(red)(4x)sqrt(t)color(red)(dx)=`
where I multiplyed and divided by `4`:
substituting the red part with the derived bit we get:
`int1/4sqrt(t)dt=`
integrating:
`1/4intt^(1/2)dt=1/4t^(1/2+1)/(1/2+1)+c=1/4t^(3/2)/(3/2)+c=`
`=1/6tsqrt(t)+c`

but: `2x^2+7=t`

so we get:
`intxsqrt(2x^2+7)dx=1/6(2x^2+7)sqrt(2x^2+7)+c`