Q.2 (A) (iii) Evaluate : ∫frac1+xx+e^(-x) dx

Answer: log|xe^x+1|+c Multiply numerator and denominator by e^x : I=∫e^x(1+x)/xe^x+1 dx Put xe^x+1=t . Then (xe^x+e^x) dx=dt , i.e. e^x(1+x) dx=dt . I=∫dt/t=log|t|+c=log|xe^x+1|+c .

Evaluate : \(\int\frac{1+x}{x+e^{-x}}\,dx\)

Q.2 (A) (iii) Evaluate : \(\int\frac{1+x}{x+e^{-x}}\,dx\)

Doubtly Asked 28 Oct 2026 Board exam · March 2026 hsc 12th Save WhatsApp Report

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Answer: \(\log\left|xe^x+1\right|+c\)

Multiply numerator and denominator by \(e^x\):

\(I=\int\frac{e^x(1+x)}{xe^x+1}\,dx\)

Put \(xe^x+1=t\). Then \((xe^x+e^x)\,dx=dt\), i.e. \(e^x(1+x)\,dx=dt\).

\(I=\int\frac{dt}{t}=\log|t|+c=\log\left|xe^x+1\right|+c\).

Doubtly AI · 28 Oct 2026 AI-verified · Oct 2026 Report

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Evaluate : \(\int\frac{1+x}{x+e^{-x}}\,dx\)

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