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3.3 · Q2

Q.Evaluate the following:

(i) ∫ex[x−2−2x−3] dx\int e^x[x^{-2}-2x^{-3}]\,dx
(ii) ∫e3x[log⁡3x+13x]dx\int e^{3x}\left[\log 3x+\frac{1}{3x}\right]dx
(iii) ∫e2x[2x−14x2]dx\int e^{2x}\left[\frac{2x-1}{4x^2}\right]dx
(iv) ∫[log⁡log⁡x+1(log⁡x)2]dx\int \left[\log\log x+\frac{1}{(\log x)^2}\right]dx
Puducherry CbseNCERTSubjective· 5mImportance★★★★★
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Each integrand fits ∫eax[af(x)+f′(x)]dx=eaxf(x)+C\int e^{ax}\big[af(x)+f'(x)\big]dx=e^{ax}f(x)+C; spot ff and read off the answer.

Standard result: ∫eax[af(x)+f′(x)]dx=eaxf(x)+C\displaystyle\int e^{ax}\big[af(x)+f'(x)\big]dx=e^{ax}f(x)+C. For a=1a=1: ∫ex[f(x)+f′(x)]dx=exf(x)+C\int e^{x}\big[f(x)+f'(x)\big]dx=e^{x}f(x)+C.

(i) ∫ex[x−2−2x−3]dx\displaystyle\int e^{x}\big[x^{-2}-2x^{-3}\big]dx

  1. Take f(x)=x−2f(x)=x^{-2}; then f′(x)=−2x−3f'(x)=-2x^{-3}, so f+f′=x−2−2x−3f+f'=x^{-2}-2x^{-3}.
  2. =ex x−2+C=exx2+C=e^{x}\,x^{-2}+C=\dfrac{e^{x}}{x^{2}}+C.

(ii) ∫e3x[log⁡3x+13x]dx\displaystyle\int e^{3x}\Big[\log 3x+\dfrac{1}{3x}\Big]dx (a=3a=3)

  1. Take f(x)=13log⁡3xf(x)=\tfrac13\log 3x; then f′(x)=13⋅1x=13xf'(x)=\tfrac13\cdot\tfrac1x=\tfrac{1}{3x} and 3f=log⁡3x3f=\log 3x, so 3f+f′=log⁡3x+13x3f+f'=\log 3x+\tfrac{1}{3x}.
  2. =e3xf(x)+C=13e3xlog⁡3x+C=e^{3x}f(x)+C=\tfrac13e^{3x}\log 3x+C.

(iii) ∫e2x[2x−14x2]dx\displaystyle\int e^{2x}\Big[\dfrac{2x-1}{4x^2}\Big]dx (a=2a=2)

  1. Simplify: 2x−14x2=12x−14x2\dfrac{2x-1}{4x^2}=\dfrac{1}{2x}-\dfrac{1}{4x^2}.
  2. Take f(x)=14xf(x)=\dfrac{1}{4x}; then f′(x)=−14x2f'(x)=-\dfrac{1}{4x^2} and 2f=12x2f=\dfrac{1}{2x}, so 2f+f′=12x−14x22f+f'=\dfrac{1}{2x}-\dfrac{1}{4x^2}.
  3. =e2xf(x)+C=e2x4x+C=e^{2x}f(x)+C=\dfrac{e^{2x}}{4x}+C.

(iv) ∫[log⁡log⁡x+1(log⁡x)2]dx\displaystyle\int\Big[\log\log x+\dfrac{1}{(\log x)^2}\Big]dx …

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