   Chapter 6.2, Problem 4QY ### 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
1 views

# In Exercises 1–6, use integration by parts to find the indefinite integral. ∫ x x + 6 3   d x

To determine

To calculate: The solution of indefinite integral xx+63dx by the use of integration by parts.

Explanation

Given Information:

The provided expression is, xx+63dx.

Formula used:

The formula for integration by parts is,

udv=uvvdu

Here, u and v be the differentiable functions x.

General Power Integral Rule:

undu=un+1n+1+C

Calculation:

Consider the provided expression,

xx+63dx

Rewrite.

xx+63dx=x(x+6)1/3dx

Let u=x and dv=(x+6)1/3dx. So,

du=dx

And,

v=(x+6)1/3dx=(x+6)13+113+1=34(x+6)4/3

Apply integration by parts to solve integral of xx+63dx.

xx+63dx=(x)(34(x+6)4/3)(34(x+6)4/3)(dx)=34x(x+6)

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