How do you integrate int 1/sqrt(4x^2+16x+13) using trigonometric substitution?

1 Answer

1/2\ln|2(x+2)+\sqrt{4x^2+16x+13}|+C

Explanation:

I=\int\frac{dx}{\sqrt{4x^2+16x+13}}

=\int \frac{dx}{2\sqrt{x^2+4x+\frac{13}{4}}}

=1/2\int \frac{dx}{\sqrt{(x+2)^2-\frac{3}{4}}}

Let x+2=\frac{\sqrt3}{2}\sec\theta \implies dx=\frac{\sqrt3}{2}\sec\theta\tan\theta d\theta

=1/2\int \frac{\frac{\sqrt3}{2}\sec\theta \tan\theta\ d\theta}{\sqrt{3/4\sec^2\theta-\frac{3}{4}}}

=1/2\int \frac{\sec\theta\tan\theta\ d\theta}{\tan\theta}

=1/2\int \sec\thetad\theta

=1/2\ln|\sec\theta+\tan\theta|+c

=1/2\ln|\frac{2}{\sqrt{3}}(x+2)+\sqrt{4/3(x+2)^2-1}|+c

=1/2\ln|(x+2)+\sqrt{(x+2)^2-3/4}|+c'

=1/2\ln|2(x+2)+\sqrt{4x^2+16x+13}|+C