Activity 64.
Evaluate each of the following indefinite integrals.
(a)
\(\displaystyle \int \frac{x^2}{5x^3+1} \, dx\)
Hint.
Note that \(5x^3 + 1\) and \(15x^2\) form a function-derivative pair.
Answer.
\(\displaystyle \int \frac{x^2}{5x^3+1} \, dx = \frac{1}{15} \ln(5x^3 + 1) + C\text{.}\)
Solution.
Since \(5x^3 + 1\) and \(15x^2\) form a function-derivative pair, we let \(u=5x^3+1\text{,}\) and observe that \(du=15x^2dx\text{,}\) and thus \(x^2dx=\frac{1}{15}du\text{.}\) Applying this substitution, integrating, and substituting back, \(\int \frac{x^2}{5x^3+1} \, dx = \int \frac{\frac{1}{15}du}{u} = \frac{1}{15} \ln(u) + C = \frac{1}{15} \ln(5x^3 + 1) + C\text{.}\)
(b)
\(\displaystyle \int e^x \sin(e^x) \, dx\)
Hint.
Recall that \(e^{x}\) is its own derivative.
Answer.
\(\displaystyle \int e^x \sin(e^x) \, dx = -\cos(e^x) + C\text{.}\)
Solution.
Because \(\frac{d}{dx}[e^x] = e^x\text{,}\) if we let \(u=e^x\text{,}\) it follows \(du = e^x dx\text{.}\) Substituting and integrating, \(\int e^x \sin(e^x) \, dx = \int \sin(u) \, du = -\cos(u) + C = -\cos(e^x) + C\text{.}\)
(c)
\(\displaystyle \int \frac{\cos(\sqrt{x})}{\sqrt{x}} \, dx\)
Hint.
Observe that \(x^{-1/2} = \frac{1}{\sqrt{x}}\text{.}\)
Answer.
\(\displaystyle \int \frac{\cos(\sqrt{x})}{\sqrt{x}} \, dx = 2\sin(\sqrt{x}) + C\text{.}\)
Solution.
Let \(u = \sqrt{x}\text{,}\) so that \(du = \frac{1}{2}x^{-1/2}dx = \frac{dx}{2\sqrt{x}}\text{.}\) We observe that \(\frac{dx}{\sqrt{x}} = 2 du\text{,}\) and thus \(\int \frac{\cos(\sqrt{x})}{\sqrt{x}}~dx = \int 2 \cos(u) \, du = 2\sin(u) + C = 2\sin(\sqrt{x}) + C\text{.}\)
