Science:Math Exam Resources/Courses/MATH101/April 2007/Question 03 (c)/Solution 1

First, we complete the square in our integral to get

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

Now, let $x+1=2\tan \theta$ so that $dx=2\sec ^{2}\theta$. This gives

$\int {\frac {dx}{\sqrt {(x+1)^{2}+4}}}=\int {\frac {2\sec ^{2}\theta d\theta }{\sqrt {4\tan ^{2}\theta +4}}}$

Using the trig identity $\tan ^{2}\theta +1=\sec ^{2}\theta$, we have

$\int {\frac {2\sec ^{2}\theta d\theta }{\sqrt {4\tan ^{2}\theta +4}}}=\int {\frac {2\sec ^{2}\theta d\theta }{\sqrt {4\sec ^{2}\theta }}}=\int \sec \theta d\theta$

This last integral has a clever trick. Multiply top and bottom by

${\frac {\sec \theta +\tan \theta }{\sec \theta +\tan \theta }}$

to get

${\begin{aligned}\int \sec \theta d\theta &=\int \sec \theta \cdot {\frac {\sec \theta +\tan \theta }{\sec \theta +\tan \theta }}d\theta \\&=\int {\frac {\sec ^{2}\theta +\sec \theta \tan \theta }{\sec \theta +\tan \theta }}d\theta \\\end{aligned}}$

Let $u=\sec \theta +\tan \theta$ so $du=\sec ^{2}\theta +\sec \theta \tan \theta$ and so the above integral is

${\begin{aligned}\int {\frac {\sec ^{2}\theta +\sec \theta \tan \theta }{\sec \theta +\tan \theta }}d\theta &=\int {\frac {du}{u}}\\&=\ln |u|\\&=\ln |\sec \theta +\tan \theta |\end{aligned}}$

To get x back in the expression we need to draw our triangle.

(Here we used Pythagorean theorem to get the hypotenuse.) From the picture, we see that

$\cos \theta ={\frac {2}{\sqrt {(x+1)^{2}+4}}}$

and so

$\sec \theta ={\frac {\sqrt {(x+1)^{2}+4}}{2}}$

Thus,

$\int {\frac {dx}{\sqrt {x^{2}+2x+5}}}=\ln |\sec \theta +\tan \theta |=\ln \left|{\frac {\sqrt {(x+1)^{2}+4}}{2}}+{\frac {x+1}{2}}\right|$

completing the question.