TS EAMCET · Maths · Indefinite Integration
\(\int \frac{x^2 \operatorname{Tan}^{-1} x}{\left(1+x^2\right)^2} d x=\)
- A \(\frac{\left(\operatorname{Tan}^{-1} x\right)^2}{4}-\frac{x \operatorname{Tan}^{-1} x}{2\left(1+x^2\right)}+\frac{1-x^2}{4\left(1+x^2\right)}+c\)
- B \(\frac{\left(\operatorname{Tan}^{-1} x\right)^2}{4}-\frac{4 x \operatorname{Tan}^{-1} x+1-x^2}{8\left(1+x^2\right)}+\mathrm{c}\)
- C \(\frac{\left(\operatorname{Tan}^{-1} x\right)^2}{4}-\frac{x \operatorname{Tan}^{-1} x}{\left(1+x^2\right)}-\frac{1-x^2}{4\left(1+x^2\right)}+\mathrm{c}\)
- D \(\frac{(\tan x)^2}{4}+\frac{4 x \operatorname{Tan}^{-1} x-1+x^2}{4\left(1+x^2\right)}+\mathrm{c}\)
Answer & Solution
Correct Answer
(B) \(\frac{\left(\operatorname{Tan}^{-1} x\right)^2}{4}-\frac{4 x \operatorname{Tan}^{-1} x+1-x^2}{8\left(1+x^2\right)}+\mathrm{c}\)
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