# integral question

• Jan 5th 2010, 05:27 PM
james121515
integral question
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
• Jan 5th 2010, 06:12 PM
Jhevon
Quote:

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

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 :confused: