How about this one...any help around?

int dx/(sqrt(2x^2+5)

1 Answer
Aug 12, 2017

int 1/(sqrt(2x^2+5)) \ dx = 1/sqrt(2) \ sinh^(-1)(sqrt(2)/sqrt(5)x) + C

Explanation:

We have:

I = int 1/(sqrt(2x^2+5)) \ dx

We can simplify and standardise the quadratic in the denominator.

Let u=sqrt(2)/sqrt(5)x => (du)/dx = sqrt(2)/sqrt(5)

Substituting into the integral we get:

I = int 1/(sqrt(2(sqrt(5)/sqrt(2)u)^2+5) )\ (sqrt(5)/sqrt(2)) \ du
\ \ = sqrt(5)/sqrt(2) \ int 1/( sqrt(5u^2+5) ) \ du
\ \ = sqrt(5)/sqrt(2) \ int 1/( sqrt(5)sqrt(u^2+1) ) \ du
\ \ = 1/sqrt(2) \ int 1/( sqrt(u^2+1) ) \ du

This is now a standard integral, so we have:

I = 1/sqrt(2) \ sinh^(-1)u + C

And, restoring the substitution, we get:

I = 1/sqrt(2) \ sinh^(-1)(sqrt(2)/sqrt(5)x) + C