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Evaluate:

\(\rm \int_1^4{\ln x\over x}dx\)


1. \(\rm 3\over4\)
2. \(\rm 1\over 4\)
3. \(\rm {(\ln 4)^2\over 2}\)
4. \(\rm \ln 4\)

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Correct Answer - Option 3 : \(\rm {(\ln 4)^2\over 2}\)

Concept:

Integral property:

  • ∫ xn dx = \(\rm x^{n+1}\over n+1\)+ C ; n ≠ -1
  • \(\rm∫ {1\over x} dx = \ln x\) + C
  • ∫ edx = ex+ C
  • ∫ adx = (ax/ln a) + C ; a > 0,  a ≠ 1
  • ∫ sin x dx = - cos x + C
  • ∫ cos x dx = sin x + C

Substitution method: If the function cannot be integrated directly substitution method is used. To integration by substitution is used in the following steps:

  • A new variable is to be chosen, say “t”
  • The value of dt is to be determined.
  • Substitution is done and the integral function is then integrated.
  • Finally, the initial variable t, to be returned.

 

Calculation:

I = \(\rm \int{\ln x\over x}dx\)

Let ln x = t ⇒\(\rm1\over x\)dx = dt

I = \(\rm \int{t}dt\)

I = \(\rm {t^2\over 2}\)

I = \(\rm {(\ln x)^2}\)

Putting the limits 

I = \(\frac{1}{2} \times \rm \left[(\ln x)^2\right]_1^4\)

I = \(\frac{1}{2} \times \rm \left[(\ln 4)^2- (\ln 1)^2\right]\)

I = \(\frac{1}{2} \times \rm (\ln 4)^2 = {({ln \ 4})^2\over 2} \)

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