Question

How do you integrate `int x/sqrt(x+1)` using substitution?

Answer 1

`y=(2(x+1)^(3/2))/3-2sqrt(x+1)+c`

Explanation:

so let us take,

`u=x+1`

so we differentiate this to find that

`(du)/(dx)=1`

and `x=u-1`

so let's rewrite our original integral

`int(x/sqrt(x+1))dx`

`=int((u-1)/sqrt(u))(1)dx`

`=int(u/sqrt(u)-1/sqrt(u))(du)/(dx)dx`

`=int(sqrt(u)-1/sqrt(u))(du)`

`=(2u^(3/2))/3-2sqrt(u)+c`

we can now sub in the `x+1 " for the u's"` if we need,

`=(2(x+1)^(3/2))/3-2sqrt(x+1)+c`

Answer 2

The integral is `=2/3sqrt(x+1)(x-2)+C`

Explanation:

Let `u=x+1` and `x=u-1`

the `du=dx`
So`int(xdx)/sqrt(x+1)=int((u-1)du)/u^(1/2)`

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

`=u^(3/2)/(3/2)-u^(1/2)/(1/2)`

`=2/3u*sqrtu-2sqrtu`
`=2*sqrtu(1/3u-1)`
`=2/3sqrtu(u-3)`

Replacing u by (x+1)

`int(xdx)/sqrt(x+1)=2/3sqrt(x+1)(x+1-3)+C`

`=2/3sqrt(x+1)(x-2)+C`