Question

How to verify by differentiation that `int1/(sqrt(x^2-a^2)dx` =`lnabs(x+sqrt(x^2-a^2)` +c, given that x>`absa` ?

Answer 1

With `x > absa` we know `abs(x+sqrt(x^2-a^2)) = x+sqrt(x^2-a^2)`, so

Explanation:

`lnabs(x+sqrt(x^2-a^2)) = ln(x+sqrt(x^2-a^2))`

We also know that `d/dx(lnu) = 1/u (du)/dx`.

And `d/dx(sqrtu) = 1/(2sqrtu) (du)/dx`

Therefore the derivative of `ln(x+sqrt(x^2-a^2))` is

`1/(x+sqrt(x^2+a^2)) * (1+1/(2sqrt(x^2-a^2)) * (2x))`

`= 1/(x+sqrt(x^2+a^2)) * (1+(2x)/(2sqrt(x^2-a^2)))`

`= 1/(x+sqrt(x^2+a^2)) * (1+x/sqrt(x^2-a^2))`

`= 1/(x+sqrt(x^2+a^2)) * (sqrt(x^2+a^2)/(sqrt(x^2+a^2))+x/sqrt(x^2-a^2))`

`= 1/sqrt(x^2+a^2)`