Question

If $\int \frac{3x+1}{\left(x-3\right)\left(x-5\right)}dx$
$= \int\frac{-5}{\left(x-3\right)}dx+\int \frac{B}{\left(x-5\right)}dx,$ then the value of $B$ is

• A. 3
• B. 4
• C. 6
• D. 8

Answer 1

Answer: (D)

We have,
$ \frac{3x+1}{\left(x-3\right)\left(x-5\right)}= \frac{-5}{x-3}+ \frac{B}{x-5}$
$3x + 1 = -5\left(x - 5\right) + B\left(x - 3\right)$
Put $x = 5$
$3\left(5\right) + 1 = B\left(5 - 3\right)$
$16 = 2B$ or $B = 8$