Question

Evaluate : $\int\frac{x^{3}+x}{x^{4}-9}dx$

• A. $\frac{1}{4}log \left|x^{4}-9\right| +\frac{1}{12}log\left|\frac{x^{2}+3}{x^{2}-3}\right| +C$
• B. $\frac{1}{4}log \left|x^{4}-9\right| -\frac{1}{12}log\left|\frac{x^{2}-3}{x^{2}+3}\right| +C$
• C. $\frac{1}{4}log \left|x^{4}-9\right| +\frac{1}{12}log\left|\frac{x^{2}-3}{x^{2}+3}\right| +C$
• D. None of these

Answer 1

Answer: (C)

$\frac{1}{4}log \left|x^{4}-9\right| +\frac{1}{12}log\left|\frac{x^{2}-3}{x^{2}+3}\right| +C$

Answer 2

Let $I = \int\frac{x^{3}+x}{x^{4}-9} dx$ $\Rightarrow I = \int \frac{x^{3}}{x^{4}-9} dx + \int\frac{x}{x^{4}-9}dx=I_{1}+I_{2}+C\left(say\right)$, where $I_{1} =\int \frac{x^{3}}{x^{4}-9}dx$ and $I_{2} = \int \frac{x}{x^{4}-9}dx$ putting $x^{4}-9=t$ in $I_{1} \Rightarrow4x^{3}dx=dt$ , we get $I_{1} = \frac{1}{4}\int\frac{1}{t}dt = \frac{1}{4}log \left|x^{4}-9\right|$ Now,$I_{2}=\int\frac{x}{x^{4}-9}dx = \int \frac{x}{\left(x^{2}\right)^{2}-3^{2}}dx$ putting $x^{2}=t \Rightarrow 2x dx = dt$, we get $I_{2}=\frac{1}{2}\int\frac{dt}{t^{2}-3^{2}} = \frac{1}{2}\cdot\frac{1}{2\times3} log\left|\frac{t-3}{t+3}\right| = \frac{1}{12} log \left|\frac{x^{2}-3}{x^{2}+3}\right|$ Hence, $I= \frac{1}{4}log \left|x^{4}-9\right| +\frac{1}{12}log\left|\frac{x^{2}-3}{x^{2}+3}\right| +C$