MHT CET · Maths · Indefinite Integration
\(\int \frac{d x}{e^x+e^{-x}+2}=\)
- A \(\frac{1}{\mathrm{e}^{2 \mathrm{x}}+1}+\mathrm{c}\)
- B \(\frac{-1}{\mathrm{e}^{\mathrm{x}}+1}+\mathrm{c}\)
- C \(\frac{1}{\mathrm{e}^{\mathrm{x}}}+\mathrm{c}\)
- D \(\frac{-1}{\mathrm{e}^{\mathrm{x}}}+\mathrm{c}\)
Answer & Solution
Correct Answer
(B) \(\frac{-1}{\mathrm{e}^{\mathrm{x}}+1}+\mathrm{c}\)
Step-by-step Solution
Detailed explanation
Let
\(\begin{aligned}
I & =\int \frac{d x}{e^x+e^{-x}+2} \\
& =\int \frac{d x}{e^x+\frac{1}{e^x}+2}=\int \frac{e^x d x}{e^{2 x}+2 e^x+1}=\int \frac{e^x}{\left(e^x+1\right)^2} d x
\end{aligned}\)
Put \(\mathrm{e}^{\mathrm{x}}+1=\mathrm{t} \Rightarrow \mathrm{e}^{\mathrm{x}} \mathrm{dx}=\mathrm{dt}\)
\(I=\int \frac{d t}{t^2}=\int t^2 d t=\frac{t^{-1}}{-1}+c=\frac{-1}{t}+c=\frac{-1}{e^x+1}+c\)
\(\begin{aligned}
I & =\int \frac{d x}{e^x+e^{-x}+2} \\
& =\int \frac{d x}{e^x+\frac{1}{e^x}+2}=\int \frac{e^x d x}{e^{2 x}+2 e^x+1}=\int \frac{e^x}{\left(e^x+1\right)^2} d x
\end{aligned}\)
Put \(\mathrm{e}^{\mathrm{x}}+1=\mathrm{t} \Rightarrow \mathrm{e}^{\mathrm{x}} \mathrm{dx}=\mathrm{dt}\)
\(I=\int \frac{d t}{t^2}=\int t^2 d t=\frac{t^{-1}}{-1}+c=\frac{-1}{t}+c=\frac{-1}{e^x+1}+c\)
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