MHT CET · Maths · Indefinite Integration
\(\int \frac{x^{e-1}+e^{x-1}}{x^{e}+e^{x}} d x\) is equal to
- A \(\log \left(x^{e}+e^{x}\right)+c\)
- B \(e \log \left(x^{e}+e^{x}\right)+c\)
- C \(\frac{1}{e} \log \left(x^{e}+e^{x}\right)+c\)
- D None of the above
Answer & Solution
Correct Answer
(C) \(\frac{1}{e} \log \left(x^{e}+e^{x}\right)+c\)
Step-by-step Solution
Detailed explanation
Put
\(x^{e}+e^{x}=t \)
\( \Rightarrow e\left(x^{e-1}+e^{x-1}\right) d x =d t \)
\( \therefore \int \frac{x^{e-1}+e^{x-1}}{x^{e}+e^{x}} d x =\frac{1}{e} \int \frac{d t}{t} \)
\( =\frac{1}{e} \log t+c \)
\( =\frac{1}{e} \log \left(x^{e}+e^{x}\right)+c\)
\(x^{e}+e^{x}=t \)
\( \Rightarrow e\left(x^{e-1}+e^{x-1}\right) d x =d t \)
\( \therefore \int \frac{x^{e-1}+e^{x-1}}{x^{e}+e^{x}} d x =\frac{1}{e} \int \frac{d t}{t} \)
\( =\frac{1}{e} \log t+c \)
\( =\frac{1}{e} \log \left(x^{e}+e^{x}\right)+c\)
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