Question

How do you find the Integral of `dx/sqrt(x^2+16)`?

Answer 1

`int (dx)/ sqrt(x^2+16) dx = int 1/(4sqrt((x/4)^2+1)) dx`

`=int 1/(sqrt((x/4)^2+1))( 1/4)dx`

`u = x/4`, `du=1/4 dx` and `int 1/sqrt(u^2+1) du=sinh^(-1) u +C`

So
`int (dx)/ sqrt(x^2+16) dx = sinh^(-1) (x/4) +C`