Question

What is the antiderivative of `F(x) = x/ sqrt (x^2 + 1)`?

Answer 1

`sqrt(x^2+1)+C`

Explanation:

Writing your integral in the form

`1/2*int2x/sqrt(x^2+1)dx`

Now we Substitute

`x^2+1=t`
then

`2xdx=dt`

and our integral is
`1/2int dt/sqrt(t)`
`1/2*int t^{-1/2}dt=t^(1/2)+C`
So we obtain

`sqrt(x^2+1)+C`

Answer 2

Note: `color(blue)(int(f'(x))/sqrt(f(x))dx=2sqrt(f(x))+c`

`I=intx/sqrt(x^2+1)dx=1/2int(2x)/sqrt(x^2+1)dx=1/2xx2sqrt(x^2+1)+c`

`:.I=sqrt(x^2+1)+c`

Explanation:

`II^(nd) Method :`

`I=intx/sqrt(x^2+1)dx`

Let,

`sqrt(x^2+1)=u=>x^2+1=u^2=>2xdx=2udu=>xdx=udu`

So,

`I=intu/sqrt(u^2)du=intu/udu=int1du=u+c`

Subst. `u=sqrt(x^2+1)`

`=>I=sqrt(x^2+1)+c`