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Worked Examples · Example 10

Q.Evaluate:

(a) ∫ex[x2+2x] dx\int e^x[x^2+2x]\,dx
(b) ∫ex[log⁡x+1x]dx\int e^x\left[\log x+\frac{1}{x}\right]dx
(c) ∫ex[x−1x2]dx\int e^x\left[\frac{x-1}{x^2}\right]dx
(d) ∫[1log⁡x−1(log⁡x)2]dx\int \left[\frac{1}{\log x}-\frac{1}{(\log x)^2}\right]dx
Puducherry CbseNCERTSubjective· 5mImportance★★★★★
17% · 10/59 Questions
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Parts (a)–(c) fit the pattern ∫ex[f(x)+f′(x)]dx=exf(x)+C\int e^{x}\big[f(x)+f'(x)\big]dx=e^{x}f(x)+C; part (d) fits ∫[ϕ′(x)]dx\int\big[\phi'(x)\big]dx-style recognition with ϕ=x/log⁡x\phi=x/\log x.

∫ex[f(x)+f′(x)]dx=exf(x)+C.\int e^{x}\big[f(x)+f'(x)\big]dx=e^{x}f(x)+C.

Steps

  1. (a) Write x2+2x=f+f′x^2+2x=f+f' with f=x2f=x^2, f′=2xf'=2x. So ∫ex(x2+2x) dx=exx2+C=x2ex+C.\int e^{x}(x^2+2x)\,dx=e^{x}x^2+C=x^2e^{x}+C.

  2. (b) With f=log⁡xf=\log x, f′=1xf'=\dfrac1x: ∫ex(log⁡x+1x)dx=exlog⁡x+C.\int e^{x}\big(\log x+\tfrac1x\big)dx=e^{x}\log x+C.

  3. (c) x−1x2=1x−1x2\dfrac{x-1}{x^2}=\dfrac1x-\dfrac{1}{x^2}. With f=1xf=\dfrac1x, f′=−1x2f'=-\dfrac{1}{x^2}: ∫ex(1x−1x2)dx=exx+C.\int e^{x}\big(\tfrac1x-\tfrac{1}{x^2}\big)dx=\dfrac{e^{x}}{x}+C.

  4. (d) Try ϕ(x)=xlog⁡x\phi(x)=\dfrac{x}{\log x}. Then …

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