Question

How do you find the integral `intx^2/((x^2+2)^(3/2))dx` ?

Answer 1

`=1/2*x/sqrt(2+x^2)+c`, where `c` is a constant

Explanation :

`=intx^2/((x^2+2)^(3/2))dx`

`=intx^2/(x^3(1+2/x^2)^(3/2))dx`

`=int1/(x(1+2/x^2)^(3/2))dx`

Using Integration by Substitution,

let's assume `2/x^2=t`

then `-4/xdx=dt`

`=int-dt/(4(1+t)^(3/2))`

`=-1/4int(1+t)^(-3/2)dt`

`=-1/4*((1+t)^(-1/2))/(-1/2)`

`=1/2*1/sqrt(1+t)+c`, where `c` is a constant

Substituting `t` back,

`=1/2*1/sqrt(1+2/x^2)+c`, where `c` is a constant

`=1/2*x/sqrt(2+x^2)+c`, where `c` is a constant