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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
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The correct answer is:
8
We have,
$$
\begin{array}{l}
\frac{3 x+1}{(x-3)(x-5)}=\frac{-5}{x-3}+\frac{B}{x-5} \\
3 x+1=-5(x-5)+B(x-3) \\
\text { Put } x=5 \\
3(5)+1=B(5-3) \\
16=2 B \text { or } B=8
\end{array}
$$
$$
\begin{array}{l}
\frac{3 x+1}{(x-3)(x-5)}=\frac{-5}{x-3}+\frac{B}{x-5} \\
3 x+1=-5(x-5)+B(x-3) \\
\text { Put } x=5 \\
3(5)+1=B(5-3) \\
16=2 B \text { or } B=8
\end{array}
$$
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