Question

Use integration by parts twice to find the exact value of integral $\int_{0}^{e} x (\ln x)^{2} d x$

Accepted Answer

Let $I = \int_{0}^{e} x {(\ln x)}^{2} d x$

Consider, $\int x \cdot {(\ln x)}^{2} d x$

Let us perform integration by parts by taking ${(\ln x)}^{2}$ as first function and $x$ as second function.

$∴ \int x \cdot {(\ln x)}^{2} d x = {(\ln x)}^{2} \int x d x - \int \{\frac{d}{d x} {(\ln x)}^{2} \int x d x\} d x$

$= {(\ln x)}^{2} \cdot \frac{x^{2}}{2} - \int (\frac{2 \ln x}{x} \cdot \frac{x^{2}}{2}) d x$

$= \frac{x^{2} {(\ln x)}^{2}}{2} - \int x \ln x d x$

Again perform integration by parts in $\int x \ln x d x$ by taking $\ln x$ as first function and $x$ as second function.

$∴ \frac{x^{2} {(\ln x)}^{2}}{2} - \int x \ln x d x = \frac{x^{2} {(\ln x)}^{2}}{2} - [\ln x \int x d x - \int \{\frac{d}{d x} (\ln x) \int x d x\} d x]$

$= \frac{x^{2} {(\ln x)}^{2}}{2} - [\frac{x^{2} \ln x}{2} - \int (\frac{1}{x} \times \frac{x^{2}}{2}) d x]$

$= \frac{x^{2} {(\ln x)}^{2}}{2} - [\frac{x^{2} \ln x}{2} - \frac{1}{2} \int x d x]$

$= \frac{x^{2} {(\ln x)}^{2}}{2} - [\frac{x^{2} \ln x}{2} - \frac{x^{2}}{4}] + C$

$= \frac{x^{2} {(\ln x)}^{2}}{2} - \frac{x^{2} \ln x}{2} + \frac{x^{2}}{4} + C$

Hence, $I = {[\frac{x^{2} {(\ln x)}^{2}}{2} - \frac{x^{2} \ln x}{2} + \frac{x^{2}}{4}]}_{0}^{e}$

$= \frac{e^{2} {(\ln (e))}^{2}}{2} - \frac{e^{2} (\ln (e))}{2} + \frac{e^{2}}{4} - 0 + 0 - 0$

$= \frac{e^{2}}{2} - \frac{e^{2}}{2} + \frac{e^{2}}{4}$

$= \frac{e^{2}}{4}$

Sue Pemberton, Julianne Hughes and, Julian Gilbey Solutions for Chapter: Further Calculus, Exercise 7: EXERCISE 8F

Sue Pemberton Mathematics Solutions for Exercise - Sue Pemberton, Julianne Hughes and, Julian Gilbey Solutions for Chapter: Further Calculus, Exercise 7: EXERCISE 8F

Questions from Sue Pemberton, Julianne Hughes and, Julian Gilbey Solutions for Chapter: Further Calculus, Exercise 7: EXERCISE 8F with Hints & Solutions

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