Question

$\int \frac{dx}{x(x-2)(x-4)}=$

• A. $\frac{1}{8}\log |x|-\frac{1}{4}\log |x-2|-\frac{1}{8}\log |x-4|+c$
• B. $\frac{1}{8}\log |x|-\frac{1}{4}\log |x-2|+\frac{1}{8}\log |x-4|+c$
• C. $\frac{1}{8}\log |x|+\frac{1}{4}\log |x-2|-\frac{1}{8}\log |x-4|+c$
• D. $\frac{1}{8}\log |x|+\frac{1}{4}\log |x-2|+\frac{1}{8}\log |x-4|+c$

Answer 1

The correct option is B $\frac{1}{8}\log |x|-\frac{1}{4}\log |x-2|+\frac{1}{8}\log |x-4|+c$

$I=\frac{1}{x\left(x-2\right)\left(x-4\right)}=\frac{A}{x}+\frac{B}{x-2}+\frac{C}{x-4}\Rightarrow \frac{1}{\left(-2\right)\left(-4\right)}=A\Rightarrow A=\frac{1}{8}\Rightarrow \frac{1}{2\left(-2\right)}=B\Rightarrow B=-\frac{1}{4}\Rightarrow \frac{1}{4\left(4-2\right)}=C\Rightarrow C=\frac{1}{8}now,\Rightarrow \frac{1}{8}{\int }^{}\frac{dx}{x}-\frac{1}{4}{\int }^{}\frac{dx}{x-2}+\frac{1}{8}{\int }^{}\frac{dx}{x-4}\Rightarrow \frac{1}{8}\log \left|x\right|-\frac{1}{4}\log \left|x-2\right|+\frac{1}{8}\log \left|x-4\right|+C$

Hence, option $B$ is correct answer,