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

x(x+1)1/3dx

Short Answer

Expert verified

The solution of the given integral is x(x+1)1/3dx=37(x+1)7/3-32(x+1)4/3+C.

Step by step solution

01

Step 1. Given Information 

Solving the given integrals.

x(x+1)1/3dx

02

Step 2. Using the substitution method.

Let

u=x+1dudx=1du=dx

In this question in which integration by substitution works even though the integrand is not at all in the form f'(u(x))u'(x). A clever change of variables will allow us to rewrite the integral so that it can be algebraically simplified.

x=u-1

03

Step 3. This substitution changes the integral into 

x(x+1)1/3dx=(u-1)u1/3dux(x+1)1/3dx=(uu1/3-1u1/3)dux(x+1)1/3dx=(u1+1/3-u1/3)dux(x+1)1/3dx=(u4/3-u1/3)du

04

Step 4. After simplifying. 

x(x+1)1/3dx=u4/3du-u1/3dux(x+1)1/3dx=u4/3+14/3+1-u1/3+11/3+1+Cx(x+1)1/3dx=u7/37/3-u4/34/3+Cx(x+1)1/3dx=37u7/3-32u4/3+Cx(x+1)1/3dx=37(x+1)7/3-32(x+1)4/3+C

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