Question

Find partial fraction :

$\frac{42-19x}{(x^2+1)(x-4)}=$

• A. $\frac{2x-11}{x^2+1}+\frac{2}{x-4}$
• B. $\frac{2x+11}{x^2+1}-\frac{2}{x+4}$
• C. $\frac{2x-11}{x^2+1}-\frac{2}{x-4}$
• D. $\frac{2x+11}{x^2+1}+\frac{2}{x+4}$

Answer 1

The correct option is C $\frac{2x-11}{x^2+1}-\frac{2}{x-4}$

$\frac{42-19x}{(x^2+1)(x-4)}=\frac{A}{(x-4)}+\frac{(Bx+C)}{(x^2+1)}$

finding the values of A,B,C.

$\Rightarrow 42-19x=A(x^2+1)+(Bx+C)(x-4)$

Comparing co-efficients of

$x^2$ we get

$0=A+B--\left(1\right)$

$x$ we get

$-19=-4B+C--\left(2\right)$

1 we get

$42=A-4C--\left(3\right)$

So, it is

solving the equations

we get

$A=-2$

$B=2$

$C=-11$

$\frac{42-19x}{(x^2+1)(x-4)}=\frac{-2}{x-4}+\frac{2x-11}{x^2+1}$