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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.)

1x1+3x2dx

Short Answer

Expert verified

The solution of the given integral is 1x1+3x2dx=13tan-13x16ln1+3x2+C.

Step by step solution

01

Step 1. Given Information 

Solving the given integrals.

1x1+3x2dx

02

Step 2. Solving the given integral using substitution method. 

Let

u=1+3x2v=3xdudx=6xdvdx=3du=6xdxdv=3dx16du=xdx13dv=dx

03

Step 3. This substitution changes the integral into 

1x1+3x2dx=11+3x2x1+3x2dx1x1+3x2dx=11+3x2dxx1+3x2dx1x1+3x2dx=1311+v2dv161udu1x1+3x2dx=13tan-1v16lnu+C1x1+3x2dx=13tan-13x16ln1+3x2+C

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