Question

Resolve into partial fractions : $\frac{26x^2+208x}{(x^2+1)(x+5)}$

• A. $\frac{41x+3}{x^2+1}-\frac{15}{x+5}$
• B. $\frac{4x+3}{x^2-1}-\frac{5}{x+5}$
• C. $\frac{-41x+3}{x^2+1}+\frac{15}{x+5}$
• D. $\frac{x-3}{x^2-1}-\frac{5}{x-5}$

Answer 1

The correct option is A $\frac{41x+3}{x^2+1}-\frac{15}{x+5}$

The first thing is to factor the denominator and get the form of the partial fraction decomposition

$\Rightarrow$ $\frac{26x^2+208x}{(x^2+1)(x+5)}=\frac{Ax+B}{x^2+1}+\frac{C}{x+5}$

The next step is to set numerators equal

$\Rightarrow$$26x^2+208x=(Ax+B)(x+5)+C(x^2+1)$

$\Rightarrow$$26x^2+208x=(A+C)x^2-x(5A+B)+(5B+C)$

we can just pick a few values of $x$ and find the constants so let’s do that

$\Rightarrow$ $x=0$ $5B+C=0$.....................$\left(1\right)$

$\Rightarrow$ $x=1$ $2C+6A+6B=234$...........$\left(2\right)$

$\Rightarrow$ $x=-1$ $2C-4A+4B=-182$......$\left(3\right)$

Now, by solving eq $1,2,3$, we get

$A=41,$ $B=3,$ $C=-15$

$\Rightarrow$ $\frac{26x^2+208}{(x^2+1)(x+5)}=\frac{41x+3}{x^2+1}-\frac{15}{x+5}$