Q.1 (B) (ii) For ∫(x-1)/((x+1)³)e^x dx=e^x· f(x)+c, where f(x)=(x+1)².

Answer: False (x-1)/((x+1)³)=((x+1)-2)/((x+1)³)=1/((x+1)²)+(-2)/((x+1)³). With f(x)=1/((x+1)²), f'(x)=(-2)/((x+1)³), so the integral is e^x[f(x)+f'(x)] form and equals e^x/((x+1)²)+c. Hence f(x)=(x+1)⁻², not (x+1)².

Chapter: Integration· Mathematics & Statistics (Commerce) Part 1

For \(\int\frac{x-1}{(x+1)^3}e^x\,dx=e^x\cdot f(x)+c\), where \(f(x)=(x+1)^2\).

Q.1 (B) (ii) For \(\int\frac{x-1}{(x+1)^3}e^x\,dx=e^x\cdot f(x)+c\), where \(f(x)=(x+1)^2\).

Doubtly Asked 29 Oct 2026 Board exam · July 2025 hsc 12th Save WhatsApp Report

1 answer

Best answer

Answer: False

\(\frac{x-1}{(x+1)^3}=\frac{(x+1)-2}{(x+1)^3}=\frac{1}{(x+1)^2}+\frac{-2}{(x+1)^3}\). With \(f(x)=\frac{1}{(x+1)^2}\), \(f'(x)=\frac{-2}{(x+1)^3}\), so the integral is \(e^x[f(x)+f'(x)]\) form and equals \(\frac{e^x}{(x+1)^2}+c\). Hence \(f(x)=(x+1)^{-2}\), not \((x+1)^2\).

Doubtly AI · 29 Oct 2026 AI-verified · Nov 2026 Report

Was this answer helpful?

For \(\int\frac{x-1}{(x+1)^3}e^x\,dx=e^x\cdot f(x)+c\), where \(f(x)=(x+1)^2\).

1 answer

Your answer