$\text { Use partial fractions to find } \int \frac { x ^ { 2 } + x + 5 } { x ^ { 4 } + 10 x ^ { 2 } + 25 } d x \text {. }$
A)
$\int \frac { x ^ { 2 } + x + 5 } { x ^ { 4 } + 10 x ^ { 2 } + 25 } d x = ( 1 / \sqrt { 5 } ) \arctan ( x / \sqrt { 5 } ) - \frac { 1 } { 2 \left( x ^ { 2 } + 5 \right) } + C$
B)
$\int \frac { x ^ { 2 } + x + 5 } { x ^ { 4 } + 10 x ^ { 2 } + 25 } d x = \frac { 1 } { 2 \left( x ^ { 2 } + 10 \right) } + C$
C)
$\int \frac { x ^ { 2 } + x + 5 } { x ^ { 4 } + 10 x ^ { 2 } + 25 } d x = ( 1 / \sqrt { 5 } ) \arcsin ( x / \sqrt { 5 } ) + C$
D)
$\int \frac { x ^ { 2 } + x + 5 } { x ^ { 4 } + 10 x ^ { 2 } + 25 } d x = - \frac { 1 } { 2 \left( x ^ { 2 } + 5 \right) } + C$
E)
$\int \frac { x ^ { 2 } + x + 5 } { x ^ { 4 } + 10 x ^ { 2 } + 25 } d x = \arctan ( x / \sqrt { 10 } ) - \frac { 1 } { 2 \left( x ^ { 2 } + 5 \right) } + C$

Question

$\text { Use partial fractions to find } \int \frac { x ^ { 2 } + x + 5 } { x ^ { 4 } + 10 x ^ { 2 } + 25 } d x \text {. }$ 
A)
$\int \frac { x ^ { 2 } + x + 5 } { x ^ { 4 } + 10 x ^ { 2 } + 25 } d x = ( 1 / \sqrt { 5 } ) \arctan ( x / \sqrt { 5 } ) - \frac { 1 } { 2 \left( x ^ { 2 } + 5 \right) } + C$
B)
$\int \frac { x ^ { 2 } + x + 5 } { x ^ { 4 } + 10 x ^ { 2 } + 25 } d x = \frac { 1 } { 2 \left( x ^ { 2 } + 10 \right) } + C$
C)
$\int \frac { x ^ { 2 } + x + 5 } { x ^ { 4 } + 10 x ^ { 2 } + 25 } d x = ( 1 / \sqrt { 5 } ) \arcsin ( x / \sqrt { 5 } ) + C$
D)
$\int \frac { x ^ { 2 } + x + 5 } { x ^ { 4 } + 10 x ^ { 2 } + 25 } d x = - \frac { 1 } { 2 \left( x ^ { 2 } + 5 \right) } + C$
E)
$\int \frac { x ^ { 2 } + x + 5 } { x ^ { 4 } + 10 x ^ { 2 } + 25 } d x = \arctan ( x / \sqrt { 10 } ) - \frac { 1 } { 2 \left( x ^ { 2 } + 5 \right) } + C$

Quizplus · Accepted Answer

First we factor the denominator: 
$$x^4+10x^2+25=(x^2+5)^2$$
Next, we find the partial fraction decomposition: 
$$\frac{x^2+x+5}{x^4+10x^2+25}=\frac{Ax+B}{x^2+5}+\frac{Cx+D}{(x^2+5)^2}$$
Multiplying both sides by $(x^2+5)^2$ and setting $x=0,1,-\sqrt{5},\sqrt{5}$ leads to the equations: 
\begin{align*} B&=1 \ A+D&=0 \ 2A+C&=1 \ 2D-C&=5 \end{align*}
Solving these equations gives $A=-\frac{1}{10}$, $B=1$, $C=\frac{1}{10}$, and $D=\frac{1}{2}$. Therefore, 
\begin{align*} \int \frac{x^2+x+5}{x^4+10x^2+25}dx &=\int\left(-\frac{1}{10}\cdot \frac{x}{x^2+5}+\frac{1}{x^2+5}-\frac{1}{10}\cdot \frac{x}{(x^2+5)^2}ight)dx \ &=-\frac{1}{10}\ln\left(x^2+5ight)+\frac{1}{\sqrt{5}}\arctan\left(\frac{x}{\sqrt{5}}ight)-\frac{1}{10}\cdot\frac{1}{2}\left(\frac{2x}{x^2+5}ight)+C \ &=-\frac{1}{10}\ln\left(x^2+5ight)+\frac{1}{\sqrt{5}}\arctan\left(\frac{x}{\sqrt{5}}ight)-\frac{1}{2\left(x^2+5ight)}+C \end{align*} Therefore, the best choice is $\boxed{	extbf{(A)}}$.

Use partial fractions to find ∫x2+x+5x4+10x2+25dx. \text { Use partial fractions to find } \int \frac { x ^ { 2 } + x + 5 } { x ^ { 4 } + 10 x ^ { 2 } + 25 } d x \text {. } Use partial fractions to find ∫x4+10x2+25x2+x+5​dx.

$\text { Use partial fractions to find } \int \frac { x ^ { 2 } + x + 5 } { x ^ { 4 } + 10 x ^ { 2 } + 25 } d x \text {. }$