MCQ
Evaluate: $\int \frac{x^3+x}{x^4-9} d x$
- 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
- 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
Evaluate: $\int \frac{x^3+x}{x^4-9} d x$
(c) : Let $I=\int \frac{x^3+x}{x^4-9} d x$
$=\int \frac{x^3}{x^4-9} d x+\int \frac{x}{x^4-9} d x=I_1+I_2+C$ (say), where
$I_1=\int \frac{x^3}{x^4-9} d x$ and $I_2=\int \frac{x}{x^4-9} d x$
Putting $x^4-9=t$ in $I_1$
$\Rightarrow 4 x^3 d x=d t$, we get
$I_1=\frac{1}{4} \int \frac{1}{t} d t=\frac{1}{4} \log |t|=\frac{1}{4} \log \left|x^4-9\right|$
Now, $I_2=\int \frac{x}{x^4-9} d x=\int \frac{x}{\left(x^2\right)^2-3^2} d x$
Putting $x^2=t \Rightarrow 2 x d x=d t$, we get
$I_2=\frac{1}{2} \int \frac{d t}{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, $I=\frac{1}{4} \log \left|x^4-9\right|+\frac{1}{12} \log \left|\frac{x^2-3}{x^2+3}\right|+C$
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