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Q.(a) Evaluate : ∫01xex(x+1)2 dx\int_0^1 \dfrac{xe^x}{(x+1)^2}\,dx

(OR)
(b) Solve the following differential equation : dydx=ex+y+x2ey\dfrac{dy}{dx} = e^{x+y} + x^2 e^y
CBSECBSE Class XII Board 2022Subjective· 2mImportance★★★★★
✓ Free question

  1. Use the ∫ex[f(x)+f′(x)] dx=exf(x)+C\int e^x[f(x)+f'(x)]\,dx = e^x f(x)+C identity to get e2−1\frac{e}{2}-1.
  2. Separate variables to get ex+e−y+x33=Ce^x + e^{-y} + \frac{x^3}{3} = C.

  1. ∫ex[f(x)+f′(x)] dx=exf(x)+C\displaystyle\int e^x\big[f(x)+f'(x)\big]\,dx = e^x f(x)+C, where f′(x)f'(x) is the derivative of f(x)f(x).
  2. Variable-separable form dydx=g(x)h(y)⇒∫dyh(y)=∫g(x) dx\dfrac{dy}{dx}=g(x)h(y)\Rightarrow \displaystyle\int\dfrac{dy}{h(y)}=\int g(x)\,dx.

Part (a): Evaluate ∫01xex(x+1)2 dx\displaystyle\int_0^1 \dfrac{xe^x}{(x+1)^2}\,dx.

  1. Split the algebraic factor by writing x=(x+1)−1x=(x+1)-1: x(x+1)2=(x+1)−1(x+1)2=1x+1−1(x+1)2\dfrac{x}{(x+1)^2}=\dfrac{(x+1)-1}{(x+1)^2}=\dfrac{1}{x+1}-\dfrac{1}{(x+1)^2}.
  2. So the integrand is ex[1x+1−1(x+1)2]e^x\left[\dfrac{1}{x+1}-\dfrac{1}{(x+1)^2}\right].
  3. Let f(x)=1x+1f(x)=\dfrac{1}{x+1}. Then f′(x)=−1(x+1)2f'(x)=-\dfrac{1}{(x+1)^2}, so the bracket is exactly f(x)+f′(x)f(x)+f'(x).
  4. By the identity, ∫xex(x+1)2 dx=exf(x)+C=exx+1+C\displaystyle\int \dfrac{xe^x}{(x+1)^2}\,dx = e^x f(x)+C = \dfrac{e^x}{x+1}+C.
  5. Apply the limits 00 to 11: [exx+1]01=e11+1−e00+1=e2−1\left[\dfrac{e^x}{x+1}\right]_0^1 = \dfrac{e^1}{1+1}-\dfrac{e^0}{0+1} = \dfrac{e}{2}-1.
  6. Numerically, e2−1=2.718282−1=1.3591−1=0.3591\dfrac{e}{2}-1 = \dfrac{2.71828}{2}-1 = 1.3591-1 = 0.3591.

Part (b): Solve dydx=ex+y+x2ey\dfrac{dy}{dx}=e^{x+y}+x^2 e^y.

  1. Factor the right side: ex+y+x2ey=ey ex+x2ey=ey(ex+x2)e^{x+y}+x^2 e^y = e^y\,e^x + x^2 e^y = e^y\big(e^x+x^2\big).
  2. Separate variables: dyey=(ex+x2) dx\dfrac{dy}{e^y}=\big(e^x+x^2\big)\,dx, i.e. e−y dy=(ex+x2) dxe^{-y}\,dy=\big(e^x+x^2\big)\,dx.
  3. Integrate both sides: ∫e−y dy=∫(ex+x2) dx\displaystyle\int e^{-y}\,dy=\int \big(e^x+x^2\big)\,dx.
  4. This gives −e−y=ex+x33+C1-e^{-y}=e^x+\dfrac{x^3}{3}+C_1.
  5. Rearranging, ex+e−y+x33=Ce^x+e^{-y}+\dfrac{x^3}{3}=C (where C=−C1C=-C_1 is an arbitrary constant).
✓Final answer

  1. ∫01xex(x+1)2 dx=e2−1≈0.359\displaystyle\int_0^1 \dfrac{xe^x}{(x+1)^2}\,dx = \dfrac{e}{2}-1 \approx 0.359
  2. General solution: ex+e−y+x33=Ce^x + e^{-y} + \dfrac{x^3}{3} = C

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