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

Question

If $ \int \frac{3 x+1}{(x-3)(x-5)} d x=\int \frac{-5}{(x-3)} d x+\int \frac{B}{(x-5)} d x $, then the value of $ B $ is, $ 3 $, $ 4 $, $ 6 $, $ 8 $

MARKS App · Accepted Answer

Given, ∫ 3 x + 1 x - 3 x - 5 d x = ∫ - 5 x - 3 d x + ∫ B x - 5 d x So, let 3 x + 1 x - 3 x - 5 = − 5 x − 3 + B x − 5 ⇒ 3 x + 1 = − 5 x - 5 + B x - 3 Putting x = 5 in the above equation, we get 3 5 + 1 = 0 + B 5 - 3 ⇒ 16 = 2 B or B = 8 .

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