(a)OR(b)

Find : \[\int \frac{2 \mathrm{x}}{\left(\mathrm{x}^{2}+3\right)\left(\mathrm{x}^{2}-5\right)} d \mathrm{x} \]

Let $\displaystyle I=\int \frac{2 \mathrm{x}}{\left(\mathrm{x}^{2}+3\right)\left(\mathrm{x}^{2}-5\right)} \mathrm{dx}$

\[\begin{aligned} & \text { Put } \mathrm{x}^{2}=t \Rightarrow 2 \mathrm{x} . d \mathrm{x}=d t \\ & \begin{aligned} \Rightarrow I & =\int \frac{d t}{(t+3)(t-5)} \\ = & \int\left(-\frac{1}{8(t+3)}+\frac{1}{8(t-5)}\right) d t \\ = & \frac{1}{8}[\log |t-5|-\log |t+3|]+c \\ = & \frac{1}{8} \log \left|\frac{\mathrm{x}^{2}-5}{\mathrm{x}^{2}+3}\right|+c \end{aligned} \end{aligned} \]

$\displaystyle \int_{1}^{4}(|\mathrm{x}-2|+|\mathrm{x}-4|) d \mathrm{x}$

\[\begin{aligned} & =\int_{1}^{2}(2-\mathrm{x}) d \mathrm{x}+\int_{2}^{4}(\mathrm{x}-2) d \mathrm{x}-\int_{1}^{4}(\mathrm{x}-4) d \mathrm{x} \\ & =\left[\frac{(2-\mathrm{x})^{2}}{-2}\right]_{1}^{2}+\left[\frac{(\mathrm{x}-2)^{2}}{2}\right]_{2}^{4}-\left[\frac{(\mathrm{x}-4)^{2}}{2}\right]_{1}^{4} \\ & =\frac{1}{2}+2+\frac{9}{2}=7 \end{aligned} \]