Find the indefinite integral $\int \frac { x ^ { 2 } - 25 } { x ^ { 2 } + 25 } d x$.
A) $\frac { x ^ { 3 } - 25 x } { x ^ { 3 } + 25 x } + C$
B)
$x - 10 \arctan \left( \frac { x } { 5 } \right) + C$
C)
$x - 6 \arctan \left( \frac { x } { 5 } \right) + C$
D) $\frac { x ^ { 3 } - 75 x } { x ^ { 3 } + 75 x } + C$

Question

Quizplus · Accepted Answer

We can use substitution $u = x^2+25$, then $du = 2xdx$:
$$\int \frac{x^2-25}{x^2+25}dx = \int \frac{du}{2u} = \frac{1}{2} \ln|u| + C = \frac{1}{2} \ln|x^2+25| + C$$
Using the identity $\arctan 	heta = \frac{1}{2} \ln \left| \frac{1+	heta}{1-	heta} ight|$ and substituting back for $u$:
$$\frac{1}{2} \ln|x^2+25| = \frac{1}{2} \ln \left| \frac{x^2+25}{1+\left(\frac{x}{5}ight)^2} ight| = \frac{1}{2} \ln | x^2+25 | - \frac{1}{2} \ln \left| 1+\left(\frac{x}{5}ight)^2 ight| = \frac{1}{2} \ln |x^2+25| - \arctan \left( \frac{x}{5} ight)$$
Therefore, the answer is:
$$\int \frac{x^2-25}{x^2+25}dx = x - 10 \arctan \left( \frac{x}{5} ight) + C$$

Find the Indefinite Integral ∫x2−25x2+25dx\int \frac { x ^ { 2 } - 25 } { x ^ { 2 } + 25 } d x∫x2+25x2−25​dx

Find the Indefinite Integral $\int \frac { x ^ { 2 } - 25 } { x ^ { 2 } + 25 } d x$