$\int_0^2 \frac{x-\mathrm{a}}{x+\mathrm{a}} \mathrm{d} x=$

$\int_0^2 \frac{x-\mathrm{a}}{x+\mathrm{a}} \mathrm{d} x=$
  1. $\mathrm{a}-2 \mathrm{a} \log 2$
  2. $a-a \log 2$
  3. $\mathrm{a}+2 \mathrm{a} \log 2$
  4. $a+a \log 2$

Solution

$\begin{aligned} & \text { Let } \mathrm{I}=\int_0^{\mathrm{a}} \frac{x-\mathrm{a}}{x+\mathrm{a}} \mathrm{~d} x \\ & \text { Let } x+\mathrm{a}=\mathrm{t} \\ & \Rightarrow x=\mathrm{t}-\mathrm{a} \\ & \text { If } x=0 \text {, then } \mathrm{t}=\mathrm{a} \\ & \text { If } x=\mathrm{a} \text {, then } \mathrm{t}=2 \mathrm{a} \\ & \therefore \quad \mathrm{~d} x=\mathrm{dt} \\ & \therefore \quad I=\int_a^{2 a} \frac{t-2 a}{t} d t \\ & =\int_a^{2 a} 1 d t-2 a \int_a^{2 a} \frac{1}{t} d t=[t]_a^{2 a}-2 a[\log t]_a^{2 a} \\ & =\mathrm{a}-2 \mathrm{a}(\log 2 \mathrm{a}-\log \mathrm{a}) \\ & =a-2 a \log 2 \end{aligned}$

Asked in: MHT CET 2024 (10 May Shift 2)

Practice more Definite Integration questions on Aicharya