First slide

Methods of integration

Question

$\int | x | \log | x | dx is equal to (x \neq 0)$

Moderate

A

$\frac{x^{2}}{2} \log | x | - \frac{x^{2}}{4} + C$
B

$\frac{1}{2} x | x | \log x + \frac{1}{4} x | x | + C$
C

$- \frac{x^{2}}{2} \log | x | + \frac{x^{2}}{4} + C$
D

$\frac{1}{2} x | x | \log | x | - \frac{1}{4} x | x | + C$

Solution

$Case II If x < 0, then | x | = - x$
$\begin{array}{l} ∴ \int | x | \log | x | dx = \int xlog xdx \\ = \log x \cdot \frac{x^{2}}{2} - \int \frac{1}{x} \cdot \frac{x^{2}}{2} dx \\ = \frac{x^{2}}{2} \cdot \log x - \frac{x^{2}}{4} + C = + \frac{x^{2}}{2} \log | x | - \frac{x^{2}}{4} + C \end{array}$
$Case II If x < 0, then | x | = - x$
$\begin{array}{l} ∴ \int | x | \log | x | dx = - \int xlog (- x) dx \\ = - \{\log (- x) \cdot \frac{x^{2}}{2} - \frac{x^{2}}{4}\} + C \\ = - \frac{x^{2}}{2} \log | x | + \frac{x^{2}}{4} + C \end{array}$
On combining both cases, we get
$\int | x | \log | x | dx = \frac{1}{2} x | x | \log | x | - \frac{1}{4} x | x | + C$

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