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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,
$\begin{array}{l}
\frac{3 x+1}{(x-3)(x-5)}=\frac{-5}{x-3}+\frac{\mathrm{B}}{x-5} \\
3 x+1=-5(x-5)+\mathrm{B}(x-3) \\
\text { Put } x=5 \\
3(5)+1=\mathrm{B}(5-3) \\
16=2 \mathrm{~B} \text { or } \mathrm{B}=8
\end{array}$
$\begin{array}{l}
\frac{3 x+1}{(x-3)(x-5)}=\frac{-5}{x-3}+\frac{\mathrm{B}}{x-5} \\
3 x+1=-5(x-5)+\mathrm{B}(x-3) \\
\text { Put } x=5 \\
3(5)+1=\mathrm{B}(5-3) \\
16=2 \mathrm{~B} \text { or } \mathrm{B}=8
\end{array}$
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