U-Substitution of Indefinite Integrals Examples 2
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U-Substitution of Indefinite Integrals Examples 2
We will now look at more examples of U-Substitution. If you haven't yet seen the U-Substitution Examples 1, please review them as these examples will be more difficult.
Example 1
Evaluate the following integral $\int \frac{x + 5}{2x + 3} \: dx$.
First let $u = 2x + 3$ so that $du = 2 dx$. Divide both sides by 2 to get $\frac{du}{2} = dx$. We should note that $u + 7 = 2x + 10$. Dividing both sides by 2, we get $\frac{u + 7}{2} = x + 5$. Making the appropriate substitutions:
(1)\begin{align} \int \frac{x + 5}{2x + 3} \: dx = \int \frac{1}{2} \cdot \frac{1}{2} \cdot \frac{u + 7}{u} \: du \\ \int \frac{x + 5}{2x + 3} \: dx = \frac{1}{4} \int 1 + 7u^{-1} \: du \\ \int \frac{x + 5}{2x + 3} \: dx = \frac{1}{4} (u + 7 \ln \mid u \mid ) \\ \quad \int \frac{x + 5}{2x + 3} \: dx = \frac{1}{4} [(2x + 3) + 7 \ln \mid (2x + 3) \mid ] + C \end{align}
Example 2
Evaluate the following integral $\int \frac{(3 + \ln x)^2(2 - \ln x)}{4x} \: dx$.
Let $u = 3 + \ln x$. Therefore, $du = \frac{1}{x} \: du$. We note that $-u + 5 = 3- \ln u$. Making the appropriate substitutions, we get that:
(2)\begin{align} \int \frac{(3 + \ln x)^2(2 - \ln x)}{4x} \: dx = \int \frac{u^2(-u + 5)}{4} \: du \\ \int \frac{(3 + \ln x)^2(2 - \ln x)}{4x} \: dx = \frac{1}{4} \int -u^3 + 5u^2 \: du \\ \int \frac{(3 + \ln x)^2(2 - \ln x)}{4x} \: dx = \frac{1}{4} \left ( \frac{-u^4}{4} + \frac{5u^3}{3} \right ) \\ \: \int \frac{(3 + \ln x)^2(2 - \ln x)}{4x} \: dx = \frac{1}{4} \left ( \frac{-(3 + \ln x)^4}{4} + \frac{5(3 + \ln x)^3}{3} \right ) + C \end{align}
Example 3
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