   Chapter 12.2, Problem 23E ### Mathematical Applications for the ...

11th Edition
Ronald J. Harshbarger + 1 other
ISBN: 9781305108042

#### Solutions

Chapter
Section ### Mathematical Applications for the ...

11th Edition
Ronald J. Harshbarger + 1 other
ISBN: 9781305108042
Textbook Problem

# Evaluate the integrals in Problems 7-36. Check your results by differentiation. ∫ ( x 3 + 1 ) 2 ( 3 x   d x )

To determine

To calculate: The value of the integral (x3+1)23xdx and also check the solution by differentiation.

Explanation

Given Information:

The provided integral is (x3+1)23xdx.

Formula used:

According to the power formula of integrals, if u=u(x), then,

undu=un+1n+1+C

According to the power rule of derivative,

ddx(xn)=nxn1

Calculation:

Consider the provided integral,

(x3+1)23xdx

Now, let x3+1=u

Differentiate x3+1=u with respect to x and get,

3x2dx=du

Here, this substitution does not work.

Hence, this method fails.

Now, simplify the integrand algebraically as,

(x3+1)23xdx=(x6+2x3+1)3xdx=(3x7+6x4+3x)dx

Use the power rule of integrals,

undu=un+1n+1+C

Now, integrate the above expression with respect to x as,

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