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Find the indefinite integral. (Use C for the constant of integration.) \frac{5}{\sqrt {25} - (x...

Question:

Find the indefinite integral. (Use C for the constant of integration.) {eq}\frac{5}{\sqrt {25} - (x + 5)^2}dx {/eq}

Evaluate the Integral:

Let us assume that {eq}f(x) {/eq} is the function of {eq}x {/eq} and we want to find a integral of {eq}f(x) . {/eq} To find this integral we need to find the anti-derivative of {eq}f(x) {/eq} with respect to {eq}x {/eq} using the substitution method.

Answer and Explanation:

Consider the integral

{eq}\displaystyle \int \frac{5}{\sqrt {25} - (x + 5)^2}dx\\ \displaystyle =5\cdot \int \frac{1}{5-\left(x+5\right)^2}dx\\ \displaystyle =5\cdot \int \frac{1}{-\left(x+5\right)^2+5}dx {/eq}

Let us assume that

{eq}\displaystyle u=x+5\Rightarrow du=dx\\ \displaystyle =5\cdot \int \frac{1}{-u^2+5}du {/eq}

Let us assume that

{eq}\displaystyle u=\sqrt{5}v\Rightarrow du=u=\sqrt{5}dv\\ \displaystyle =5\cdot \int \frac{1}{\sqrt{5}\left(-v^2+1\right)}dv\\ \displaystyle =5\cdot \frac{1}{\sqrt{5}}\cdot \int \frac{1}{-v^2+1}dv\\ \displaystyle =5\cdot \frac{1}{\sqrt{5}}\left(\frac{\ln \left|v+1\right|}{2}-\frac{\ln \left|v-1\right|}{2}\right),\quad \left [ \int \frac{1}{-v^2+1}dv=\frac{\ln \left|v+1\right|}{2}-\frac{\ln \left|v-1\right|}{2} \right ] {/eq}

Substitute back {eq}\displaystyle v=\frac{u}{\sqrt{5}},\:u=x+5 {/eq}

{eq}\displaystyle =5\cdot \frac{1}{\sqrt{5}}\left(\frac{\ln \left|\frac{x+5}{\sqrt{5}}+1\right|}{2}-\frac{\ln \left|\frac{x+5}{\sqrt{5}}-1\right|}{2}\right)\\ \displaystyle =\sqrt{5}\left(\frac{1}{2}\ln \left|\frac{x+5}{\sqrt{5}}+1\right|-\frac{1}{2}\ln \left|\frac{x+5}{\sqrt{5}}-1\right|\right)+C {/eq}


Learn more about this topic:

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Evaluating Definite Integrals Using the Fundamental Theorem

from AP Calculus AB: Exam Prep

Chapter 16 / Lesson 2
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