Q.1 (B)(ii) State whether true or false: For displaystyle∫ (x-1)/((x+1)³) e^x dx = e^x f(x) + c,\ f(x) = (x+1)²

Answer: False Write (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 of the form ∫ e^x[f(x) + f'(x)] dx = e^x f(x) + c. Hence f(x) = 1/((x+1)²) = (x+1)⁻², not (x+1)².

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

State whether true or false: For \(\displaystyle\int \frac{x-1}{(x+1)^3}\,e^x\,dx = e^x f(x) + c,\ f(x) = (x+1)^2\)

Q.1 (B)(ii) State whether true or false: For \(\displaystyle\int \frac{x-1}{(x+1)^3}\,e^x\,dx = e^x f(x) + c,\ f(x) = (x+1)^2\)

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Answer: False

Write \(\dfrac{x-1}{(x+1)^3} = \dfrac{(x+1) - 2}{(x+1)^3} = \dfrac{1}{(x+1)^2} - \dfrac{2}{(x+1)^3}\). With \(f(x) = \dfrac{1}{(x+1)^2}\), \(f'(x) = -\dfrac{2}{(x+1)^3}\), so the integral is of the form \(\int e^x[f(x) + f'(x)]\,dx = e^x f(x) + c\). Hence \(f(x) = \dfrac{1}{(x+1)^2} = (x+1)^{-2}\), not \((x+1)^2\).

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State whether true or false: For \(\displaystyle\int \frac{x-1}{(x+1)^3}\,e^x\,dx = e^x f(x) + c,\ f(x) = (x+1)^2\)

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