# Evaluate: $\rm \int_1^4{\ln x\over x}dx$

12 views
in Calculus
closed

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. $\rm \ln 4$

by (30.0k points)
selected

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