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Chapter 5: Q. 43 (page 417)

Solve each of the integrals in Exercises 21鈥�70. Some integrals require substitution, and some do not. (Exercise 69 involves a hyperbolic function.)

鈭�lnxxdx

Short Answer

Expert verified

The solution of the given integral is 鈭�lnxxdx=23(lnx)3/2+C.

Step by step solution

01

Step 1. Given Information 

Solving the given integrals.

鈭�lnxxdx

02

Step 2. Solving the given integral using substitution method. 

Let

u=lnxdudx=1xdu=1xdx

03

Step 3. This substitution changes the integral into 

鈭�lnxxdx=鈭�udu鈭�lnxxdx=鈭�u1/2du鈭�lnxxdx=u1/2+11/2+1+C鈭�lnxxdx=u3/23/2+C鈭�lnxxdx=23u3/2+C鈭�lnxxdx=23(lnx)3/2+C

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Most popular questions from this chapter

Solve given definite integral.

鈭�35鈥�x2鈭�9dx

Solve given definite integral.

鈭�1/41/2鈥�1x21鈭�x2dx

Explain why, if x=asinu, then a2鈭�x2=acosu. Your explanation should include a discussion of domains and absolute values.

Explain why, if x=asecu, then a2sec2u鈭�a2is 鈭�atanuif x<鈭�aand is atanuif x>a. Your explanation should include a discussion of domains and absolute values.

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.

鈭�x3-3x2+2x-3x2+1dx
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