Discussion Forum

1)

If $\int_{}^{}\frac{3x + 1}{(x-5)(x-3)}$dx =

$\int_{}^{}\frac{-5}{(x-3)}dx + \int_{}^{}  \frac{B}{x-5}dx$, then the value of B is  

Answer:

Option D

Explanation:

We have,

$\frac{ 3x +1}{(x-3)(x-5)}$ = $\frac{ -5}{(x-3)}+\frac{B}{(x-5)}$

3x+1 = -5(x-5) + B(x-3)

Put x = 5

3(5)+1 + B(5-3)

16 = 2B or B = 8