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 $$\displaystyle (adjacent)^2 + (opposite)^2 = (hypotenuse)^2 \ \ \longrightarrow$$ $$(1)^2 + (x+2)^2 = (hypotenuse)^2 \ \ \longrightarrow \ \ \ hypotenuse = \sqrt{x^2+4x+5} \ \ \longrightarrow$$ $$\sec \theta = \displaystyle{ hypotenuse \over adjacent }= \displaystyle{ \sqrt{x^2+4x+5} \over 1 = \sqrt{x^2+4x+5} } . \Big)$$ $$= \displaystyle { \sqrt{x^2+4x+5} - 2 \ln \Big| \sqrt{x^2+4x+5} + (x+2) \Big| + C }$$