Question

Use Exercise 35 to find \(\int {{{(\ln x)}^3}dx} \).

Short answer

The value of \(\int {{{(\ln x)}^3}} dx\) is \(x((\ln x) - 3{(\ln x)^2} + 6(\ln x) - 6) + c\)

Step-by-step solution

Step-1: Given Equation

Given data is \(\int {{{(\ln x)}^3}dx} \)

Step-2: Evaluating the integration

\(\int {{{(\ln x)}^3}dx} \)

Let \(\ln x = t \Rightarrow x = {e^t}\)

Differentiating with respect to ‘\(x\)’

\(\frac{{dx}}{{dt}} = {e^t}\)

Step-3: Finding \(\int {{t^3}{e^t}dt} \) by using integration by parts:

\(\int {udv = uv - \int {vdu \to (2)} } \)

Let \(u = {t^3}\)

Differentiating with respect to ‘\(x\)’

\(\frac{{du}}{{dt}} = 3{t^2}\)

Let \(u = {t^3}\)

Differentiating with respect to ‘t’

\(\frac{{du}}{{dt}} = 3{t^2}\)

And \(dv = {e^t}dt\)

\(\int {dv = \int {{e^t}dt} } \)

Substituting the above values in Equation (2)

Step-4: Finding \(\int {{e^t}{t^2}dt} \) by integration parts:

\(\int {udv = uv - \int {vdu} } \)

Let \(u = {t^2}\)

Differentiating with respect to ‘t’

Step-5: Finding \(\int {{e^t}tdt} \) by using integration parts:

Let \(u = t\)

Differentiating with respect to ‘t’

and

\(\therefore \int {{e^t}tdt = t({e^t}) - \int {{e^t}dt} } \)

\( = t{e^t} - {e^t}\)

Step-6: Substituting Equation (5) in Equation (4):

Substituting value \(\int {{t^2}} {e^t}dt\) in Equation (2)

Step-7: Substituting value of \(\int {{t^3}{e^t}} dt\) in Equation (1)

\(\therefore \int {{{(\ln x)}^3}dx} = {t^3}{e^t} - 3{t^2}{e^t} + 6t{e^t} - 6{e^t} + c\) where c is integrating constant

\(\int {{{(\ln x)}^3}dx} = {e^t}({t^3} - 3{t^2} + 6t - 6) + c\)

From step 2, \(t = \ln x\)

Hence the answer for the given integration is

\(\int {{{(\ln x)}^3}dx} = x({(\ln x)^3} - 3{(\ln x)^2} + 6(\ln x) - 6) + c\)