U-Substitution with Exponential & Logarithmic Functions
Core Concepts
Focuses on applying u-substitution to integrals involving the natural exponential function (e^x) and those that result in the natural logarithm (ln|u|).
U-Substitution with Exponential & Logarithmic Functions
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What general form of an integral, after substitution, results in a natural logarithm function?
An integral of the form \(\int \frac{1}{u} \, du\) or \(\int u^{-1} \, du\). Its solution is \(\ln|u| + C\).
For the integral \(\int \frac{2x}{x^2 + 5} \, dx\), what is the correct choice for \(u\) to get a logarithmic result?
Let \(u\) be the denominator, \(u = x^2 + 5\). Its derivative is \(du = 2x \, dx\), which is exactly the numerator.
Solve \(\int \frac{2x}{x^2 + 5} \, dx\).
With \(u = x^2 + 5\) and \(du = 2x \, dx\), the integral becomes \(\int \frac{1}{u} \, du = \ln|u| + C\). Back-substituting gives \(\ln|x^2 + 5| + C\). Since \(x^2+5\) is always positive, this can be written as \(\ln(x^2 + 5) + C\).
How would you find \(\int \frac{1}{ax+b} \, dx\)?
Let \(u = ax+b\), so \(du = a \, dx\) and \(dx = \frac{1}{a}du\). The integral becomes \(\int \frac{1}{u} \cdot \frac{1}{a}du = \frac{1}{a} \int \frac{1}{u}du = \frac{1}{a} \ln|u| + C\). The solution is \(\frac{1}{a} \ln|ax+b| + C\).
What is the integral of the natural exponential function, \(\int e^u \, du\)?
$$e^u + C$$
For the integral \(\int x^2 e^{x^3} \, dx\), what is the appropriate u-substitution?
Let \(u\) be the exponent of \(e\). Set \(u = x^3\), which gives \(du = 3x^2 \, dx\).
Solve \(\int x^2 e^{x^3} \, dx\).
Let \(u=x^3\), so \(du=3x^2\,dx\) and \(x^2\,dx = \frac{1}{3}du\). The integral is \(\int e^u \cdot \frac{1}{3}du = \frac{1}{3} \int e^u du = \frac{1}{3}e^u + C\). Back-substituting gives \(\frac{1}{3}e^{x^3} + C\).
What is the best choice for \(u\) in the integral \(\int \frac{\ln(x)}{x} \, dx\)?
Let \(u = \ln(x)\). Its derivative is \(du = \frac{1}{x} \, dx\), which is also present in the integral.
Solve \(\int \frac{\ln(x)}{x} \, dx\).
With \(u = \ln(x)\) and \(du = \frac{1}{x} \, dx\), the integral becomes \(\int u \, du = \frac{u^2}{2} + C\). Back-substituting gives \(\frac{(\ln(x))^2}{2} + C\).
What is the correct substitution for \(\int \frac{e^x}{1+e^x} \, dx\)?
Let \(u\) be the entire denominator, \(u = 1+e^x\). This works because the derivative, \(du = e^x \, dx\), is the numerator.
Solve \(\int \frac{e^x}{1+e^x} \, dx\).
With \(u=1+e^x\) and \(du=e^x\,dx\), the integral is \(\int \frac{1}{u}du = \ln|u|+C = \ln|1+e^x|+C\). Since \(1+e^x > 0\), the answer is \(\ln(1+e^x)+C\).
