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Chapter 6: Problem 10

Use integration tables to find the integral. $$ \int \frac{1}{x^{2}+2 x+2} d x $$

Short Answer

Expert verified
\[\int \frac{1}{x^2+2x+2} dx = arctan(x + 1) + C \]

Step by step solution

01

Simplify Denominator

First, simplify the denominator by completing the square. The general form is \(x^2 + bx +c\) which simplifies to \((x+b/2)^2 + (c - (b^2 / 4))\). Here, \(b = 2\) and \(c = 2\), so this simplifies the denominator to: \((x + 1)^2 + 1\). So, the integral becomes, \(\int \frac{1}{(x+1)^2 + 1} dx\).
02

Comparing with a Known Integral Form

This is starting to look like the integral of a function we can find in an integral table. Specifically, the arctan function. The integral \(\int \frac{1}{1 + x^2} dx = arctan(x) + C\). To fit the integral into this form, denote \(x + 1\) as \(u\), then apply the substitution method. The differential \(du = dx\). The new integral becomes, \(\int \frac{1}{u^2 + 1} du\)
03

Evaluate the Integral Using the Table

Now, we can use the integral table to find the integral of the new integral: \(\int \frac{1}{u^2 + 1} du = arctan(u) + C\).
04

Substitute u Back to Get x

We made a substitution for \(x + 1 = u\), so we now substitute \(u\) back in terms of \(x\). The result is: \(\int \frac{1}{x^2+2x+2} dx = arctan(x + 1) + C\)

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