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MHT CET · Maths · Indefinite Integration

\(\int \frac{\mathrm{e}^{\tan ^{-1} x}}{1+x^2}\left[\left(\sec ^{-1} \sqrt{1+x^2}\right)^2+\cos ^{-1}\left(\frac{1-x^2}{1+x^2}\right)\right] \mathrm{d} x, x>0=\)

  1. A \(\left(\tan ^{-1} x\right)^2 \mathrm{e}^{\tan ^{-1} x}+\mathrm{c}\), where \(\mathrm{c}\) is a constant of integration.
  2. B \(\left(\tan ^{-1} x\right) \mathrm{e}^{\tan ^{-1} x}+\mathrm{c}\), where \(\mathrm{c}\) is a constant of integration.
  3. C \(\left(\tan ^{-1} x\right) \mathrm{e}^{2 \tan ^{-1} x}+\mathrm{c}\), where \(\mathrm{c}\) is a constant of integration.
  4. D \(\left(\tan ^{-1} x\right)^2 \mathrm{e}^{2 \tan ^{-1} x}+\mathrm{c}\), where \(\mathrm{c}\) is a constant of integration.
Verified Solution

Answer & Solution

Correct Answer

(A) \(\left(\tan ^{-1} x\right)^2 \mathrm{e}^{\tan ^{-1} x}+\mathrm{c}\), where \(\mathrm{c}\) is a constant of integration.

Step-by-step Solution

Detailed explanation

\(\int \frac{\mathrm{e}^{\tan ^{-1} x}}{1+x^2}\left[\left(\sec ^{-1} \sqrt{1+x^2}\right)^2+\cos ^{-1}\left(\frac{1-x^2}{1+x^2}\right)\right] \mathrm{d} x, \)
\( \text {Put } x=\tan \mathrm{t} \)
\( \therefore \mathrm{d} x=\sec ^2 \mathrm{t} d \mathrm{t}\)
\(\therefore \mathrm{I} =\int \frac{\mathrm{e}^{\tan ^{-1}(\tan \mathrm{t})}}{1+\tan ^2 \mathrm{t}}[\left(\sec ^{-1} \sqrt{1+\tan ^2} \mathrm{t}\right)^2+\cos ^{-1}\) \(\left(\frac{1-\tan ^2 \mathrm{t}}{1+\tan ^2 \mathrm{t}}\right)] \)
\( =\int \frac{\mathrm{e}^{\mathrm{t}}}{\sec ^2 \mathrm{t}}\left[\left(\sec ^2 \mathrm{t}(\sec \mathrm{t})\right)^2+\cos ^{-1}(\cos 2 \mathrm{t})\right] \sec ^2 \mathrm{t} d \mathrm{t} \)
\( =\int \mathrm{e}^{\mathrm{t}}\left[\mathrm{t}^2+2 \mathrm{t}\right] \mathrm{dt} \)
\( =\mathrm{e}^{\mathrm{t}} \cdot \mathrm{t}^2+\mathrm{c} \quad \ldots\left[\int \mathrm{e}^x \mathrm{f}(x) \cdot \mathrm{f}^{\prime}(x)=\mathrm{e}^x \mathrm{f}(x)+\mathrm{c}\right] \)
\( =\mathrm{t}^2 \cdot \mathrm{e}^{\mathrm{t}}+\mathrm{c} \quad \)
\( =\left(\tan ^{-1} \cdot x\right)^2 \mathrm{e}^{\tan \mathrm{tan}^{-1} x}+\mathrm{c}\)