Integral Calculus
86307
\(\int \frac{\mathrm{e}^{\tan ^{-1} \mathrm{x}}}{1+\mathrm{x}^{2}}\left[\left(\sec ^{-1} \sqrt{1+\mathrm{x}^{2}}\right)^{2}+\cos ^{-1}\left(\frac{1-\mathrm{x}^{2}}{1+\mathrm{x}^{2}}\right)\right] \mathrm{dx}=\)
1 \(\mathrm{e}^{\tan -1_{x}}\left(\tan ^{-1} \mathrm{x}\right)^{2}+\mathrm{C}\)
2 \(\mathrm{e}^{\tan -1_{x}}\left(\sec ^{-1} \mathrm{x}\right)^{2}+\mathrm{C}\)
3 \(\mathrm{e}^{\tan -1 \mathrm{x}}\left(\sec ^{-1}\left(\sqrt{1+\mathrm{x}^{2}}\right)\right)+\mathrm{C}\)
4 \(\mathrm{e}^{\tan -1_{x}}\left(\cos ^{-1}\left(\frac{1-\mathrm{x}^{2}}{1+\mathrm{x}^{2}}\right)\right)+\mathrm{C}\)
Explanation:
(A) : Given,
\(I=\int \frac{e^{\tan ^{-1} x}}{1+x^{2}}\left[\left(\sec ^{-1} \sqrt{1+x^{2}}\right)^{2}+\cos ^{-1}\left(\frac{1-x^{2}}{1+x^{2}}\right)\right] d x\)
Let, \(\quad \sec ^{-1} \sqrt{1+x^{2}}=\tan ^{-1} x\)
\(\cos ^{-1}\left(\frac{1-x^{2}}{1+x^{2}}\right)=2 \tan ^{-1} x\)
\(I=\int \frac{e^{\tan ^{-1} x}}{1+x^{2}}\left[\left(\tan ^{-1} x\right)^{2}+2 \tan ^{-1} x\right] d x\)
Let,
\(\tan ^{-1} \mathrm{x}=\mathrm{t}\)
\(\frac{1}{1+\mathrm{x}^{2}} \mathrm{dx}=\mathrm{dt}\)
\(I=\int e^{t}\left(t^{2}+2 t\right) d t\)
Here, \(\quad f(t)=t^{2} f^{\prime}(t)=2 t\)
\(\because \quad \int \mathrm{e}^{\mathrm{t}}\left(\mathrm{f}(\mathrm{t})+\mathrm{f}^{\prime}(\mathrm{t})\right) d \mathrm{t}=\mathrm{e}^{\mathrm{t}} \times \mathrm{f}(\mathrm{t})+\mathrm{C}\)
\(=\mathrm{t}^{2} \cdot \mathrm{e}^{\mathrm{t}}+\mathrm{C}\)
\(I=e^{\tan ^{-1} x}\left(\tan ^{-1} x\right)^{2}+C\)
