KCET · Maths · Indefinite Integration
The value of \(\int \frac{x^{2}+1}{x^{2}-1} d x\) is
- A \(\log \left(\frac{x-1}{x+1}\right)+c\)
- B \(\log \left(\frac{x+1}{x-1}\right)+c\)
- C \(x+\log \left(\frac{x-1}{x+1}\right)+c\)
- D \(\log \left(x^{2}-1\right)+c\)
Answer & Solution
Correct Answer
(C) \(x+\log \left(\frac{x-1}{x+1}\right)+c\)
Step-by-step Solution
Detailed explanation
Let \(I=\int \frac{x^{2}+1}{x^{2}-1} d x=\int \frac{x^{2}+1-1+1}{x^{2}-1} d x\)
\[
\begin{aligned}
&\Rightarrow I=\int \frac{x^{2}-1}{x^{2}-1} d x+\int \frac{2}{x^{2}-1} d x \\
&\Rightarrow I=\int 1 d x+2 \int \frac{1}{x^{2}-1} d x \\
&\Rightarrow I=x+2 \cdot \frac{1}{2} \log \left(\frac{x-1}{x+1}\right)+c \\
&\Rightarrow I=x+\log \left(\frac{x-1}{x+1}\right)+c
\end{aligned}
\]
\[
\begin{aligned}
&\Rightarrow I=\int \frac{x^{2}-1}{x^{2}-1} d x+\int \frac{2}{x^{2}-1} d x \\
&\Rightarrow I=\int 1 d x+2 \int \frac{1}{x^{2}-1} d x \\
&\Rightarrow I=x+2 \cdot \frac{1}{2} \log \left(\frac{x-1}{x+1}\right)+c \\
&\Rightarrow I=x+\log \left(\frac{x-1}{x+1}\right)+c
\end{aligned}
\]
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