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Chapter 4: Problem 16

Find the integral. $$ \int \frac{x-2}{(x+1)^{2}+4} d x $$

Short Answer

Expert verified
\(\frac{1}{2} \ln |(x+1)^2+4| - \frac{3}{2} \arctan \frac{x+1}{2} + C\)

Step by step solution

01

Recognize the integral

We're dealing with the integral \(\int \frac{x-2}{(x+1)^{2}+4} d x \). We should note that the derivative of \(x+1\) is \(1\), and the numerator \(x-2\) can be rewritten as \(x+1-3\), which might suggest us to perform a substitution.
02

Perform substitution

Let \(u = x + 1\), and thus \(du = dx\). Also, the integral becomes \(\int \frac{u-3}{u^2+4} du\).
03

Break down the Integral

Notice that the integral can be broken down in to two separate integrals. \(\int \frac{u-3}{u^2+4} du = \int \frac{u}{u^2+4} du - \int \frac{3}{u^2+4} du\).
04

Evaluate the integrals

For the integral of the form \(\int \frac{u}{u^2+4} du\), it evaluates to \(\frac{1}{2} \ln |u^2+4|\). For the integral of the form \(\int \frac{3}{u^2+4} du, it evaluates to \(\frac{3}{2} \arctan \frac{u}{2}\).\)
05

Back substitute the original variable

Substitute back \(u = x + 1\), we have the final answer \(\frac{1}{2} \ln |(x+1)^2+4| - \frac{3}{2} \arctan \frac{x+1}{2} + C\), where \(C\) is the constant of integration.

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