Science:Math Exam Resources/Courses/MATH103/April 2016/Question 02 (a)/Solution 1

Completing the square in the denominator, we get

$x^{2}+2x+2=(x^{2}+2x+1)+1=(x+1)^{2}+1.$

The integral can then be written as

$\int {\frac {1}{x^{2}+2x+2}}dx=\int {\frac {1}{(x+1)^{2}+1}}dx.$

We use the substitution $u=x+1,du=dx,$ to get

$\int {\frac {1}{(x+1)^{2}+1}}dx=\int {\frac {1}{u^{2}+1}}du.$

Then, we use trigonometric substitution $u=\tan \theta ,$ $du=\sec ^{2}\theta d\theta$;

$\int {\frac {1}{u^{2}+1}}du=\int {\frac {1}{\tan ^{2}\theta +1}}\cdot \sec ^{2}\theta d\theta =\int {\frac {1}{\sec ^{2}\theta }}\cdot \sec ^{2}\theta d\theta =\theta +C$.

Finally plugging $\theta =\tan ^{-1}u=\tan ^{-1}(x+1)$, we rewrite the anti-derivative in terms of $x$:

$\int {\frac {1}{x^{2}+2x+2}}dx=\theta +C=\color {blue}\tan ^{-1}(x+1)+C.$