MHT CET · Maths · Indefinite Integration
\(\int e^{\tan ^{-1} x}\left(1+\frac{x}{1+x^{2}}\right) d x\)
- A \(\left(\frac{x}{2}\right) e^{\tan ^{-1} x}+c\)
- B \(x e^{\tan ^{-1} x}+c\)
- C \(\left(\frac{1}{2}\right) e^{\tan ^{-1} x}+c\)
- D \(e^{\tan ^{-1} x}+c\)
Answer & Solution
Correct Answer
(B) \(x e^{\tan ^{-1} x}+c\)
Step-by-step Solution
Detailed explanation
Let \(I=\int e^{\tan ^{-1} x}\left(1+\frac{x}{1+x^{2}}\right) d x\)
\(I =\int e^{\tan ^{-1} x}\left(\frac{1+x^{2}+x}{1+x^{2}}\right) d x \)
\( \text { Put } \tan ^{-1} x =t \Rightarrow x=\tan t \text { and }\) \(\frac{1}{1+x^{2}} d x=d t \)
\( \therefore I =\int e^{t}\left(1+\tan ^{2} t+\tan t\right) d t \Rightarrow\) \(\int e^{t}\left(\tan t+\sec ^{2} t\right) d t \)
\( =e^{t} \tan t+c=x e^{\tan ^{-1} x}+c\)
\(I =\int e^{\tan ^{-1} x}\left(\frac{1+x^{2}+x}{1+x^{2}}\right) d x \)
\( \text { Put } \tan ^{-1} x =t \Rightarrow x=\tan t \text { and }\) \(\frac{1}{1+x^{2}} d x=d t \)
\( \therefore I =\int e^{t}\left(1+\tan ^{2} t+\tan t\right) d t \Rightarrow\) \(\int e^{t}\left(\tan t+\sec ^{2} t\right) d t \)
\( =e^{t} \tan t+c=x e^{\tan ^{-1} x}+c\)
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