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Chapter 6: Problem 32

Complete the square and find the integral. $$ \int \frac{x}{\sqrt{x^{2}-6 x+5}} d x $$

Short Answer

Expert verified
\(\frac{(x-3)^2}{2} \sqrt{(x-3)^2-4^2} - 4 ln|x-3 + \sqrt{(x-3)^2-4^2}| + 3 * ln|x-3 + \sqrt{(x-3)^2-4^2}|\)

Step by step solution

01

Complete the Square

First, complete the square for \(x^{2}-6x+5\). A perfect square trinomial is in the form \( (x-a)^2 = x^2 - 2ax + a^2 \). In this situation, \(a\) would be half of the \(x\) coefficient, which is -3. This gives \((x-3)^2\). However, when expanding \((x-3)^2\), there is a 9, not a 5. Therefore, 4 must be subtracted from the perfect square trinomial to get \(x^{2}-6x+5\). Thus, \(x^{2}-6x+5 = (x-3)^2 - 4\).
02

Insert into the Integral

Substitute the completed square into the original integral, which gives \(\int \frac{x}{\sqrt{(x^2-2*3x+3^2) - 4^2}} dx\). Now the fraction is in a form that makes substitution possible.
03

Substitution

Now, let \(u = x-3\). The differential of \(u\) is \(du = dx\). Substitute \(u\) into the integral, which gives \(\int \frac{u+3}{\sqrt{u^2 - 4^2}} du\). This integral can now be solved by integrating the functions \(\frac{1}{\sqrt{u^2-a^2}}\) whose integral is \(ln|u + \sqrt{u^2-a^2}|\) and \(u\sqrt{u^2 - a^2}\) whose integral is \(\frac{u^2}{2} \sqrt{u^2-a^2} - \frac{a^2}{2} ln|u + \sqrt{u^2-a^2}|\).
04

Integrate

This integral can now be integrated which gets \( \frac{u^2}{2} \sqrt{u^2-4^2} - 4 ln|u + \sqrt{u^2-4^2}| + 3 * ln|u + \sqrt{u^2-4^2}| \).
05

Substitute Back

Last, substitute \(x-3\) back in for \(u\), simplifying the integral to the final answer \( \frac{(x-3)^2}{2} \sqrt{(x-3)^2-4^2} - 4 ln|x-3 + \sqrt{(x-3)^2-4^2}| + 3 * ln|x-3 + \sqrt{(x-3)^2-4^2}| \). This is the solution to the integral.

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