use the accompanying sections of a table of integrals to evaluate the following indefinite integral. the…

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.\n int\frac{5e^{x}}{sqrt{25e^{2x}+9}}dx \nclick here to view basic integrals. click here to view trigonometric integrals. click here to view reduction formulas for trigonometric functions. click here to view integrals involving squares of x and a. click here to view integrals involving ax +/ - b. click here to view other integrals.\n int\frac{5e^{x}}{sqrt{25e^{2x}+9}}dx=square

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.\n int\frac{5e^{x}}{sqrt{25e^{2x}+9}}dx \nclick here to view basic integrals. click here to view trigonometric integrals. click here to view reduction formulas for trigonometric functions. click here to view integrals involving squares of x and a. click here to view integrals involving ax +/ - b. click here to view other integrals.\n int\frac{5e^{x}}{sqrt{25e^{2x}+9}}dx=square

Answer

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)