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)

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 $I_1\Rightarrow 4x^{3}dx=dt$ , 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 $x^2=t\Rightarrow 2xdx=dt$, we get

$I_2=\frac{1}{2}\int \frac{dt}{t^2-3^2}=\frac{1}{2}\cdot \frac{1}{2\times 3}log\left|\frac{t-3}{t+3}\right|=\frac{1}{12}log\left|\frac{x^2-3}{x^2+3}\right|$

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