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

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Answer 1 · 2018-06-26T18:02:03

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

#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#