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Chapter 5: Q 46. (page 429)

Solve the integral: $\int \ln \sqrt{x} d x$

Short Answer

Expert verified

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

Step by step solution

01

Step 1. Given information.

We have given integral is $\int \ln \sqrt{x} d x$ .

02

Step 2. Solve the integration by parts .

We have,

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

and

role="math" $d v = d x v = \int d x v = x$

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

$\int \ln \sqrt{x} d x = \int (\frac{1}{2} \ln x) d x = \frac{1}{2} \int \ln x d x = \frac{1}{2} (\ln (x) x - \int x . \frac{1}{x} d x) = \frac{1}{2} (\ln (x) x - \int 1 d x) = \frac{x}{2} \ln (x) - \frac{x}{2} + c$

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Most popular questions from this chapter

True/False: Determine whether each of the statements that follow is true or false. If a statement is true, explain why. If a statement is false, provide a counterexample.

(a) True or False: The substitution x = 2 sec u is a suitable choice for solving $\int \frac{1}{x^{2} - 4} d x$ .

(b) True or False: The substitution x = 2 sec u is a suitable choice for solving $\int \frac{1}{\sqrt{x^{2} - 4}} d x$ .

(c) True or False: The substitution x = 2 tan u is a suitable choice for solving $\int \frac{1}{x^{2} + 4} d x .$

(d) True or False: The substitution x = 2 sin u is a suitable choice for solving $\int {(x^{2} + 4)}^{- 5 / 2} d x$

(e) True or False: Trigonometric substitution is a useful strategy for solving any integral that involves an expression of the form $\sqrt{x^{2} - a^{2}}$ .

(f) True or False: Trigonometric substitution doesn’t solve an integral; rather, it helps you rewrite integrals as ones that are easier to solve by other methods.

(g) True or False: When using trigonometric substitution with $x = a \sin u$ , we must consider the cases $x > a$ and $x < - a$ separately.

(h) True or False: When using trigonometric substitution with $x = a s e c u$ , we must consider the cases $x > a$ and $x < - a$ separately.

Find three integrals in Exercises 27–70 for which a good strategy is to use integration by parts with $u = x$ and dv the remaining part.

Complete the square for each quadratic in Exercises 28–33. Then describe the trigonometric substitution that would be appropriate if you were solving an integral that involved that quadratic.

$x - 3 x^{2}$

Show by differentiating (and then using algebra) that $- \cot (\sin^{- 1} x)$ and $- \frac{\sqrt{1 - x^{2}}}{x}$ are both antiderivatives of $\frac{1}{x^{2} \sqrt{1 - x^{2}}}$ . How can these two very different-looking functions be an antiderivative of the same function?

Solve the integral

$\int \frac{1}{x^{2} \sqrt{x^{2} - 9}} d x$

See all solutions

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