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Integral Calculus

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Question : 57 of 73

Marks: +1, -0

Evaluate

\int x \log x d x

$x \log x + c$
$x \log x - \;\frac{x}{3} + c$
$\;\frac{x^{2}}{2} \log x - \;\frac{x^{2}}{4} + c$
$\frac{1}{x \log x} + c$

Solution:

Concept:

Use integration by parts where

\log x

as first function and

x

as second function.

Explanation:

Let

I = \int x \log x \, dx

.

Take

u = \log x

and

dv = x \, dx

.

Then

du = \frac{1}{x} dx

and

v = \frac{x^2}{2}

.

Integration by parts:

\int u \, dv = uv - \int v \, du

.

So

I = (\log x) \cdot \frac{x^2}{2} - \int \frac{x^2}{2} \cdot \frac{1}{x} dx

.

Simplify:

I = \frac{x^2}{2} \log x - \frac{1}{2} \int x \, dx

.

Evaluate:

\int x \, dx = \frac{x^2}{2}

.

Thus

I = \frac{x^2}{2} \log x - \frac{1}{2} \cdot \frac{x^2}{2} + c = \frac{x^2}{2} \log x - \frac{x^2}{4} + c

.

Answer:

\frac{x^{2}}{2} \log x - \frac{x^{2}}{4} + c

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