Use substitution and partial fractions to find the indefinite integral.
$\int \frac { \sqrt { x } } { x - 25 } d x$
A)
$\int \frac { \sqrt { x } } { x - 25 } d x = 5 \log ( 5 - \sqrt { x } ) - 4 \log ( \sqrt { x } - 5 ) + C$
B)
$\int \frac { \sqrt { x } } { x - 25 } d x = \sqrt { x } + \log ( 5 - \sqrt { x } ) - 4 \log ( \sqrt { x } - 5 ) + C$
C)
$\int \frac { \sqrt { x } } { x - 25 } d x = 2 \sqrt { x } + \log ( 1 - \sqrt { x } ) + C$
D)
$\int \frac { \sqrt { x } } { x - 25 } d x = 2 \sqrt { x } + x ^ { 2 } + 5 \log ( 5 - \sqrt { x } ) - 4 \log ( \sqrt { x } - 5 ) + C$
E)
$\int \frac { \sqrt { x } } { x - 25 } d x = 2 \sqrt { x } + 5 \log ( 5 - \sqrt { x } ) - 4 \log ( \sqrt { x } - 5 ) + C$

Question

Use substitution and partial fractions to find the indefinite integral. 
$\int \frac { \sqrt { x } } { x - 25 } d x$
A)
$\int \frac { \sqrt { x } } { x - 25 } d x = 5 \log ( 5 - \sqrt { x } ) - 4 \log ( \sqrt { x } - 5 ) + C$
B)
$\int \frac { \sqrt { x } } { x - 25 } d x = \sqrt { x } + \log ( 5 - \sqrt { x } ) - 4 \log ( \sqrt { x } - 5 ) + C$
C)
$\int \frac { \sqrt { x } } { x - 25 } d x = 2 \sqrt { x } + \log ( 1 - \sqrt { x } ) + C$
D)
$\int \frac { \sqrt { x } } { x - 25 } d x = 2 \sqrt { x } + x ^ { 2 } + 5 \log ( 5 - \sqrt { x } ) - 4 \log ( \sqrt { x } - 5 ) + C$
E)
$\int \frac { \sqrt { x } } { x - 25 } d x = 2 \sqrt { x } + 5 \log ( 5 - \sqrt { x } ) - 4 \log ( \sqrt { x } - 5 ) + C$

Quizplus · Accepted Answer

Let $u = \sqrt{x}$, then $x = u^2$ and $dx = 2u\,du$. We can substitute this into the integral to get
$$\int \frac{\sqrt{x}}{x-25}dx = \int\frac{u}{u^2-25}\cdot2u\,du = 2\int\frac{u^2}{u^2-25}\,du.$$Now we can use partial fractions to integrate this expression. We can write
$$\frac{u^2}{u^2-25} = \frac{1}{2}\left(\frac{1}{u-5}-\frac{1}{u+5}\right).$$Therefore,
$$\int \frac{\sqrt{x}}{x-25}dx = 2\int\frac{u^2}{u^2-25}\,du = 2\cdot\frac{1}{2}\int\left(\frac{1}{u-5}-\frac{1}{u+5}\right)du$$
$$= \int\frac{1}{u-5}\,du-\int\frac{1}{u+5}\,du = \ln|u-5| - \ln|u+5| + C.$$Substituting back in $u = \sqrt{x}$, we get
$$\int \frac{\sqrt{x}}{x-25}dx = \ln|\sqrt{x}-5|-\ln|\sqrt{x}+5| + C = \ln\left|\frac{\sqrt{x}-5}{\sqrt{x}+5}\right| + C.$$Using logarithmic identities, we can simplify this to
$$\ln\left|\frac{\sqrt{x}-5}{\sqrt{x}+5}\right| = \ln|\sqrt{x}-5|-\ln|\sqrt{x}+5| = \ln\left|\frac{(\sqrt{x}-5)^2}{x-25}\right|.$$Therefore,
$$\int \frac{\sqrt{x}}{x-25}dx = 2\ln\left|\frac{\sqrt{x}-5}{\sqrt{x}+5}\right| + C.$$Simplifying the logarithm using exponent rules, we get the final answer of
$$\int \frac{\sqrt{x}}{x-25}dx = \boxed{E) \; 2 \sqrt{x} + 5\ln\left|5-\sqrt{x}\right| - 4\ln\left|\sqrt{x}-5\right| + C}.$$

Use Substitution and Partial Fractions to Find the Indefinite Integral ∫xx−25dx\int \frac { \sqrt { x } } { x - 25 } d x∫x−25x​​dx

Use Substitution and Partial Fractions to Find the Indefinite Integral $\int \frac { \sqrt { x } } { x - 25 } d x$