AP EAMCET · Maths · Indefinite Integration
\(\int \frac{x^2}{\left(x^2-1\right)\left(x^2+1\right)} d x=\)
- A \(\frac{1}{4} \log \left|\frac{x+1}{x-1}\right|-\frac{1}{2} \operatorname{Tan}^{-1} x+c\)
- B \(\frac{1}{4} \log \left|\frac{x-1}{x+1}\right|+\frac{1}{2} \operatorname{Tan}^{-1} x+c\)
- C \(\frac{1}{4} \log \left|\frac{x-1}{x+1}\right|-\frac{1}{2} \operatorname{Tan}^{-1} x+c\)
- D \(\frac{1}{4} \log \left|\frac{x+1}{x-1}\right|+\frac{1}{2} \operatorname{Tan}^{-1} x+c\)
Answer & Solution
Correct Answer
(B) \(\frac{1}{4} \log \left|\frac{x-1}{x+1}\right|+\frac{1}{2} \operatorname{Tan}^{-1} x+c\)
Step-by-step Solution
Detailed explanation
Let \(\int \frac{x^2}{\left(x^2-1\right)\left(x^2+1\right)} d x=I\) \(\Rightarrow I=\frac{1}{2} \int\left(\frac{1}{x^2-1}+\frac{1}{x^2+1}\right) d x\)…
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