   Chapter 7.4, Problem 32E

Chapter
Section
Textbook Problem

# Evaluate the integral. ∫ 0 1 x x 2 + 4 x + 13 d x

To determine

To evaluate the integral 01xx2+4x+13

Explanation

Calculation: Given 01xx2+4x+13

Notice that the denominator, cannot be factored as the discriminant is less than 0.

That is, b24ac=424(1)(13)=1652<0

Since it can’t be factored, complete the square 01xx2+4x+13=01x(x+2)2+4

Let u=x+2, then du=1dx, we can write x=u2

23u2u2+9du=23uu2+92u2+9du=[12ln|u2+9|23tan−<

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