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Solve each of the integrals in Exercises 21–70. 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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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 solving1x24dx.

(b) True or False: The substitution x = 2 sec u is a suitable choice for solving1x24dx.

(c) True or False: The substitution x = 2 tan u is a suitable choice for solving1x2+4dx.

(d) True or False: The substitution x = 2 sin u is a suitable choice for solvingx2+45/2dx

(e) True or False: Trigonometric substitution is a useful strategy for solving any integral that involves an expression of the form x2a2.

(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=asinu, we must consider the cases x>a and x<-a separately.

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

Find three integrals in Exercises 39–74 that can be solved without using trigonometric substitution.

Describe two ways in which the long-division algorithm for polynomials is similar to the long-division algorithm for integers and then two ways in which the two algorithms are different.

Solve the integral:3x+1xdx

Suppose you use polynomial long division to divide p(x) by q(x), and after doing your calculations you end up with the polynomial x2-x+3 as the quotient above the top line, and the polynomial 3x − 1 at the bottom as the remainder. Thenp(x)=___andp(x)q(x)=____

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