KCET · Maths · Indefinite Integration
\(\int \mathrm{e}^{\tan ^{-1} \mathrm{x}}\left(1+\frac{\mathrm{x}}{1+\mathrm{x}^{2}}\right) \mathrm{dx}\) is equal to
- A \(x e^{\tan ^{-1} x}+c\)
- B \(\mathrm{e}^{\tan ^{-1} \mathrm{x}}+\mathrm{c}\)
- C \(\frac{1}{2} \mathrm{e}^{\tan ^{-1} \mathrm{x}}+\mathrm{c}\)
- D \(\frac{1}{2} \mathrm{xe}^{\tan ^{-1} \mathrm{x}}+\mathrm{c}\)
Answer & Solution
Correct Answer
(A) \(x e^{\tan ^{-1} x}+c\)
Step-by-step Solution
Detailed explanation
Let \(I=\int e^{\tan ^{-1} x}\left(1+\frac{x}{1+x^{2}}\right) d x\)
\(=\int e^{\tan ^{-1} x} d x+\int \frac{x e^{\tan ^{-1} x}}{1+x^{2}} d x\)
\(=x e^{\tan ^{-1} x}-\int \frac{x e^{\tan ^{-1} x}}{1+x^{2}} d x+\int \frac{x e^{\tan ^{-1} x}}{1+x^{2}} d x+c\)
\(=x e^{\tan ^{-1} x}+c\)
\(=\int e^{\tan ^{-1} x} d x+\int \frac{x e^{\tan ^{-1} x}}{1+x^{2}} d x\)
\(=x e^{\tan ^{-1} x}-\int \frac{x e^{\tan ^{-1} x}}{1+x^{2}} d x+\int \frac{x e^{\tan ^{-1} x}}{1+x^{2}} d x+c\)
\(=x e^{\tan ^{-1} x}+c\)
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