Question

What is `int (8x)/sqrt(16+x^2) dx`?

Answer 1

`8sqrt(16+x^2)+C.`

Explanation:

Let `I=int(8x)/sqrt(16+x^2)dx.`

We use Substitution `: 16+x^2=t^2.`

`:. 2xdx=2tdt.`, or, `xdx=tdt.`

Therefore, `I=int(8x)/sqrt(16+x^2)dx,=int(8t)/sqrtt^2dt=8intdt=8t=8sqrt(16+x^2)+C.`

OR

We can use the Formula : `int(f(x))^n*f'(x)dx=(f(x))^(n+1)/(n+1), n!=-1.`

Take `f(x)=(16+x^2),` `n=-1/2!=-1.` Then, `f'(x)=2x.` Thus,
`I=int(8x)/sqrt(16+x^2)dx=4int(16+x^2)^(-1/2)2xdx=4[(16+x^2)^{-1/2+1}]/(-1/2+1)=4{(16+x^2)^(1/2)}/(1/2)=8sqrt(16+x^2)+C,` as before!