$$\int \frac{\mathrm{e}^{\tan ^{-1} x}}{1+x^2}\left[\left(\sec ^{-1} \sqrt{1+x^2}\right)^2+\cos ^{-1}\left(\frac{1-x^2}{1+x^2}\right)\right] \mathrm{d} x,$$ where $x>0$ is
- A$\left(\tan ^{-1} x\right) \mathrm{e}^{\tan ^{-1} x}+\mathrm{c}$, where c is a constant of integration.
- B$\left(\tan ^{-1} x\right)^2 \mathrm{e}^{\tan ^{-1} x}+\mathrm{c}$, where c is a constant of integration.
- C$2\left(\tan ^{-1} x\right) \mathrm{e}^{\tan ^{-1} x}+\mathrm{c}$, where c is a constant of integration.
- D$2\left(\tan ^{-1} x\right)^2 \mathrm{e}^{\tan ^{-1} x}+\mathrm{c}$, where c is a constant of integration.
✓ Correct answer: B
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