Question

The value of the integral ${\int }_0^2\frac{t}{t+i}dt$ will be

• A. $2+{\tan }^{-1}2-i\frac{1}{2}\ln 5$
• B. $2-{\tan }^{-1}2+i\frac{1}{2}\ln 5$
• C. $2+{\tan }^{-1}2+i\frac{1}{2}\ln 5$
• D. $2-{\tan }^{-1}2-i\frac{1}{2}\ln 5$

Answer 1

Answer: (D)

$2-{\tan }^{-1}2-i\frac{1}{2}\ln 5$

${\int }_0^2\frac{t}{t+i}dt$

$={\int }_0^2[\frac{t}{t+i}\times \frac{t-i}{t-i}]dt$

$={\int }_0^2\frac{t^2-it}{t^2+1}dt$

$={\int }_0^2[\frac{t^2+1}{t^2+1}-\frac{1}{t^2+1}-\frac{it}{t^2+1}]dt$

$={\int }_0^{2}1dt-{\int }_0^2\frac{1}{t^2+1}dt-\frac{i}{2}{\int }_0^2\frac{2t}{t^2+1}dt$

$=[2-0]-[ta{n}^{-1}2-ta{n}^{-1}0]-\frac{i}{2}{\left[\ln \right(t^2+1\left)\right]}_0^2$

$=2-ta{n}^{-1}2-\frac{i}{2}\left[\ln \right(5)-\ln (1\left)\right]$

$=2-ta{n}^{-1}2-\frac{i}{2}\left[\ln \right(5\left)\right]$