Originally Posted by

**james121515** Hi I am working on a integral involving u substiution and inverse trig forms.

$\displaystyle

Evaluate:~~\int\frac{2}{(2x+3)\sqrt{4x^2+12x+5}}dx =

$

$\displaystyle

\int\frac{2}{(2x+3)\sqrt{\frac{x^2+3x+\frac{5}{4}} {\frac{1}{4}}}}dx

=$

Complete the square of the numberator of the radical:

$\displaystyle

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

=$

Partially expand the square-completed numerator:

$\displaystyle \int\frac{2}{(2x+3)\sqrt{\frac{(x^2+3x+\frac{9}{4} )-1}{\frac{1}{4}}}}dx

=$

Simplify the expression in the radical by elimating the denominator:

$\displaystyle \int\frac{2}{(2x+3)\sqrt{(4x^2+12x+9)-4}}dx$

Factor the portion in parenthesis in the radical:

$\displaystyle \int\frac{2}{(2x+3)\sqrt{(2x+3)^2-4}}dx$

Let $\displaystyle u = 2x + 3$

then $\displaystyle du = 2dx$.

So now we have:

$\displaystyle

\int\frac{1}{u\sqrt{u^2-4}}du

$

Notice this is the general arcsec integral formula, and so we have:

$\displaystyle

\frac{1}{2}sec^{-1}\left(\frac{|u|}{2}\right) + C

$

So substituting back in for $\displaystyle u$, we have:

$\displaystyle

\frac{1}{2}sec^{-1}\left(\frac{|2x+3|}{2}\right)+C

$

I was pretty confident this was right but when I checked it using mathematica I got:

$\displaystyle

-\frac{1}{2}tan^{-1}\frac{2}{\sqrt{(2x+3)^2-4}}$, which I checked as not being equivalent to my answer. Where might I have gone wrong?

Thanks for your help,

James