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Chapter 5: Q 29. (page 429)

Solve the integral: $\int x \ln (x) d x$

Short Answer

Expert verified

The required answer is $\frac{x^{2}}{2} \ln x - \frac{x^{2}}{4} + c$ .

Step by step solution

01

Step 1. Given information.

We have given integral is $\int x \ln (x) d x$ .

02

Step 2. Solve the integration by parts .

We have, $u = \ln x, d v = x d x$ .

$u = \ln x d u = \frac{1}{x} d x$

and

$d v = x d x v = \int x d x v = \frac{x^{2}}{2}$

The formula of integration by parts is $\int u d v = u v - \int v d u$

$\int x \ln (x) d x = \ln (x) \frac{x^{2}}{2} - \int \frac{x^{2}}{2} \times \frac{1}{x} d x = \frac{x^{2}}{2} \ln (x) - \int \frac{x}{2} d x = \frac{x^{2}}{2} \ln (x) - \frac{x^{2}}{4} + c$

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Most popular questions from this chapter

Solve the integral: $\int x \sin {(x)}^{2} d x$ .

Solve given integrals by using polynomial long division to rewrite the integrand. This is one way that you can sometimes avoid using trigonometric substitution; moreover, sometimes it works when trigonometric substitution does not apply.

$\int \frac{x^{3} - 3 x^{2} + 2 x - 3}{x^{2} + 1}$ dx

For each function u(x) in Exercises 9–12, write the differential du in terms of the differential dx.

$u (x) = \frac{1}{x}$

Explain why, if $x = a s e c u$ , then $\sqrt{a^{2} s e c^{2} u - a^{2}}$ is $- a \tan u$ if $x < - a$ and is $a \tan u$ if $x > a$ . Your explanation should include a discussion of domains and absolute values.

Write $\sin (2 \cos^{- 1} x)$ as an algebraic function.

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