   Chapter 6.2, Problem 21E ### Calculus: An Applied Approach (Min...

10th Edition
Ron Larson
ISBN: 9781305860919

#### Solutions

Chapter
Section ### Calculus: An Applied Approach (Min...

10th Edition
Ron Larson
ISBN: 9781305860919
Textbook Problem
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# Using Integration Tables In Exercises 9–36, use the integration table in Appendix C to find the indefinite integral. See Examples 1, 2,3, and 5. ∫ x 2 3 + x   d x

To determine

To calculate: The solution of indefinite integral x23+xdx.

Explanation

Given Information:

The expression is provided as, x23+xdx.

Formula used:

The integral formula is,

una+budu=2b(2n+3)[un(a+bu)3/2naun1a+budu]

Power rule of integration,

undu=un+1n+1+C

Calculation:

Consider u=x.

Differentiate the considered function with respect to x using power rule of differentiation.

du=dx

Consider the provided expression,

x23+xdx

Here, n=2, a=3, b=1, u=x and du=dx.

Substitute, n for 2, u for x, a for 3, b for 1, and du for dx.

x23+xdx=una+budu

Use the integral formula to solve the above integral as,

x23+xdx=2b(2n+3)[un(a+bu)3/2naun1a+budu]

Substitute x for u, 2 for n, 3 for a and 1 for b.

x23+xdx=2(1)(2(2)+3)[x2(3+(1)x)3/2(2)(3)x213+(1)xdx]=2(1)(4+3)[x2(3+x)3/26x3+xdx]

Further solve.

x23+xdx=27[x2(3+x)3/26x3+xdx] …… (1)

Consider the integration,

x3+xdx

Here, n=1, a=3, b=1, u=x and du=dx.

Substitute, n for 1, u for x, a for 3, b for 1, and du for dx.

x3+xdx=una+budu

By the use integral formula to solve the above integral as,

x3+xdx=2b(2n+3)[un(a+bu)3/2naun1a+budu]

Substitute x for u, 1 for n, 3 for a and 1 for b.

x3+xdx=2(1)(2(1)+3)[x1(3+(1)x)3/2(1)(3)x113+(1)xdx]=25[x(3+x)3/23(1)3+xdx]

Further solve

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