   Chapter 7.3, Problem 25E

Chapter
Section
Textbook Problem

# Evaluate the integral. ∫ x 2 3 + 2 x − x 2 d x

To determine

To Find: The value of the integral x23+2xx2dx

Explanation

Calculations: Writing 3+2xx2 as a perfect square form

3+2xx2=4(x1)2

Substitute u=x1,thendu=dx

x23+2xx2dx=(u+1)24u2du=u24u2du+2u4u2du+4u2du

From the Table of Integrals, we have u2a2u2du=u8(2u2a2)a2u2+a48sin1(ua)+C

a2u2du=u2a2u2+a22sin1(ua)+C

ua2u2du=13(a2u2)32+C

Using the formula with a = 2, we get

x23+2xx2dx=

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