Integrate:
$$\int \frac{x}{\left(x^2-4x-13\right)^2}dx$$
Here's my attempt:
I first completed the squares for the denominator:
$$\left(x^2-4x-13\right)^2=(x-2)^2-17 \implies \int \frac{x}{\left(\left(x-2\right)^2-17\right)^2}dx$$
I then used $u$-subsituition:
$$u=x-2 \implies \int \frac{u+2}{\left(u^2-17\right)^2}du = \int \frac{u}{\left(u^2-17\right)^2}du+\int \frac{2}{\left(u^2-17\right)^2}du$$
The first part of the new integral is quite simple:
$$\int \frac{u}{\left(u^2-17\right)^2}du=\frac{-1}{2(u^2-17)}$$
Then I did the second part:
$$\int \frac{2}{\left(u^2-17\right)^2}du = -\frac{1}{2\left(u^2-17\right)}+2\left(\frac{1}{68\sqrt{17}}\ln \left|u+\sqrt{17}\right|-\frac{1}{68\left(u+\sqrt{17}\right)}-\frac{1}{68\sqrt{17}}\ln \left|u-\sqrt{17}\right|-\frac{1}{68\left(u-\sqrt{17}\right)}\right) = -\frac{1}{2\left(\left(x-2\right)^2-17\right)}+2\left(\frac{1}{68\sqrt{17}}\ln \left|x-2+\sqrt{17}\right|-\frac{1}{68\left(x-2+\sqrt{17}\right)}-\frac{1}{68\sqrt{17}}\ln \left|x-2-\sqrt{17}\right|-\frac{1}{68\left(x-2-\sqrt{17}\right)}\right) = -\frac{1}{2\left(x^2-4x-13\right)}+2\left(\frac{1}{68\sqrt{17}}\ln \left|x-2+\sqrt{17}\right|-\frac{1}{68\left(x-2+\sqrt{17}\right)}-\frac{1}{68\sqrt{17}}\ln \left|x-2-\sqrt{17}\right|-\frac{1}{68\left(x-2-\sqrt{17}\right)}\right) + C, C \in \mathbb{R}$$
Is this working out correct? I'm not really sure how WolframAlpha works, so I didn't check it on there.
Best Answer
Here is an alternative method to integrate as follows $$\int \frac{x}{\left(x^2-4x-13\right)^2}dx$$ $$=\int\frac12 \frac{(2x-4)+4}{\left(x^2-4x-13\right)^2}dx$$ $$=\frac12\int \frac{2x-4}{\left(x^2-4x-13\right)^2}dx+\frac12\int\frac{4}{\left(x^2-4x-13\right)^2}dx$$ $$=\frac12\int \frac{d(x^2-4x-13)}{\left(x^2-4x-13\right)^2}+2\int\frac{d(x-2)}{\left((x-2\right)^2-17)^2}$$ using reduction formula: $\color{blue}{\int \frac{dt}{(t^2+a)^n}=\frac{t}{2(n-1)a(t^2+a)^{n-1}}+\frac{2n-3}{2(n-1)a}\int\frac{dt}{(t^2+a)^{n-1}}} $, $$=\frac12 \frac{-1}{\left(x^2-4x-13\right)}+2\left(\frac{(x-2)}{2(-17)((x-2)^2-17)}+\frac{1}{2(-17)}\int \frac{d(x-2)}{(x-2)^2-17}\right)$$ using standard formula: $\color{blue}{\int \frac{dt}{t^2-a^2}=\frac{1}{2a}\ln\left|\frac{t-a}{t+a}\right|}$, $$=-\frac{1}{2\left(x^2-4x-13\right)}-\frac{(x-2)}{17(x^2-4x-13)}-\frac{1}{34\sqrt{17}}\ln\left|\frac{x-2-\sqrt{17}}{x-2+\sqrt{17}}\right|+C $$ $$=-\frac{2x+13}{34(x^2-4x-13)}-\frac{1}{34\sqrt{17}}\ln\left|\frac{x-2-\sqrt{17}}{x-2+\sqrt{17}}\right|+C $$