1. ## integral question

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

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

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

Partially expand the square-completed numerator:
$\int\frac{2}{(2x+3)\sqrt{\frac{(x^2+3x+\frac{9}{4} )-1}{\frac{1}{4}}}}dx
=$

Simplify the expression in the radical by elimating the denominator:
$\int\frac{2}{(2x+3)\sqrt{(4x^2+12x+9)-4}}dx$

Factor the portion in parenthesis in the radical:
$\int\frac{2}{(2x+3)\sqrt{(2x+3)^2-4}}dx$

Let $u = 2x + 3$
then $du = 2dx$.

So now we have:
$
\int\frac{1}{u\sqrt{u^2-4}}du
$

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

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

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

$
-\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?
James

2. Originally Posted by james121515
Hi I am working on a integral involving u substiution and inverse trig forms.

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

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

Partially expand the square-completed numerator:
$\int\frac{2}{(2x+3)\sqrt{\frac{(x^2+3x+\frac{9}{4} )-1}{\frac{1}{4}}}}dx
=$

Simplify the expression in the radical by elimating the denominator:
$\int\frac{2}{(2x+3)\sqrt{(4x^2+12x+9)-4}}dx$

Factor the portion in parenthesis in the radical:
$\int\frac{2}{(2x+3)\sqrt{(2x+3)^2-4}}dx$

Let $u = 2x + 3$
then $du = 2dx$.

So now we have:
$
\int\frac{1}{u\sqrt{u^2-4}}du
$

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

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

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

$
-\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?
James
you made completing the square a lot harder than it should be, but what you did is fine.

i wasn't sure about the absolute values in your answer though... one of my calc texts has them, the other doesn't