Question

use the accompanying sections of a table of integrals to evaluate the following indefinite integral. the integral may require preliminary work, such as completing the squares or changing variables, before it can be found in a table.
int\frac{5e^{x}}{sqrt{25e^{2x}+9}}dx
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int\frac{5e^{x}}{sqrt{25e^{2x}+9}}dx=square

Explanation:

Step1: Let \(u = 5e^{x}\)

Differentiate \(u\) with respect to \(x\). Since \(\frac{du}{dx}=5e^{x}\), then \(du = 5e^{x}dx\). Also, \(25e^{2x}=(5e^{x})^{2}=u^{2}\). The integral \(\int\frac{5e^{x}}{\sqrt{25e^{2x}+9}}dx\) becomes \(\int\frac{du}{\sqrt{u^{2}+9}}\).

Step2: Use the integral formula

The integral formula for \(\int\frac{du}{\sqrt{u^{2}+a^{2}}}=\ln|u + \sqrt{u^{2}+a^{2}}|+C\), where \(a = 3\) (since \(a^{2}=9\)). So \(\int\frac{du}{\sqrt{u^{2}+9}}=\ln|u+\sqrt{u^{2}+9}|+C\).

Step3: Substitute back \(u = 5e^{x}\)

We get \(\ln|5e^{x}+\sqrt{25e^{2x}+9}|+C\).

Answer:

\(\ln|5e^{x}+\sqrt{25e^{2x}+9}|+C\)