SOLUTION 13: $ \ \ $ Integrate $ \displaystyle{ \int { x
\over \sqrt{x^2+4x+5} } \ dx } $. First complete the square. Then
$$ \displaystyle { \int \frac{x}{\sqrt{x^{2}+4x+5}} \ dx = \int
\frac{x}{\sqrt{(x^{2}+4x+4)+1}} \ dx} $$
$$ = \displaystyle { \int \frac{x}{\sqrt{(x+2)^{2}+1}} \ dx } $$
Use the trig substitution
$$ x+2 = \tan \theta $$
so that
$$ x = \tan \theta - 2 $$
and
$$ dx = \sec^{2} \theta \ d \theta $$
Substitute into the original problem, replacing all
forms of x, getting
$$ \displaystyle { \int \frac{x}{\sqrt{(x+2)^{2}+1}} \ dx } = \displaystyle { \int \frac{\tan \theta - 2}{\sqrt{\tan^{2}
\theta + 1}} \ \sec^{2} \theta \ d \theta } $$
$$ = \displaystyle { \int \frac{\tan \theta - 2}{\sqrt{\sec^{2}
\theta}} \ \sec^{2} \theta \ d \theta } $$
$$ = \displaystyle { \int \frac{\tan \theta - 2}{\sec \theta} \
\sec^{2} \theta \ d \theta } $$
$$ = \displaystyle { \int (\tan \theta - 2) \sec \theta \ d \theta} $$
$$ = \displaystyle { \int (\sec \theta \tan \theta - 2 \sec \theta)
\ d \theta } $$
$$ = \displaystyle { \sec \theta - 2 \ln \Big|\sec \theta + \tan
\theta \Big| + C } $$
$ \Big($ We need to write our final answer in terms of $x$.
Since $ x+2 = \tan \theta $ it follows that
$$ \tan \theta = \displaystyle{ x+2 \over 1 } = \displaystyle{ opposite \over adjacent } $$
and from the Pythagorean Theorem that
$$ \displaystyle (adjacent)^2 + (opposite)^2 = (hypotenuse)^2
\ \ \longrightarrow $$
$$ (1)^2 + (x+2)^2 = (hypotenuse)^2
\ \ \longrightarrow \ \ \ hypotenuse = \sqrt{x^2+4x+5} \ \ \longrightarrow $$
$$ \sec \theta = \displaystyle{ hypotenuse \over adjacent }= \displaystyle{ \sqrt{x^2+4x+5} \over 1
= \sqrt{x^2+4x+5} } . \Big) $$
$$ = \displaystyle { \sqrt{x^2+4x+5} - 2 \ln \Big| \sqrt{x^2+4x+5} +
(x+2) \Big| + C } $$
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