Question

1
∫ log(1+t)/1+t² dt ,Evaluate it.
0

Answer 1

Answer:

_____________

Step-by-step explanation:

___-----__________

Answer 2

$\bold{\int_{0}^{1} \frac{\log (t+1)}{t^{2}+1} d t=\frac{\pi}{8} \log 2}$

Given:

$\int_{0}^{1} \frac{\log (t+1)}{t^{2}+1} d t$

Step-by-step explanation:

Let us consider the integral as ‘I’.

$I=\int_{0}^{1} \frac{\log (t+1)}{t^{2}+1} d t$

Let $y=\tan x$

$\frac{d y}{d x}=\sec ^{2} x \Rightarrow d y=\sec ^{2} x d x$

If $x=0$ then $y=0$

If $x=1$ then $y=\frac{\pi}{4}$

$I=\int_{0}^{\frac{\pi}{4}} \frac{\log (\tan x+1)}{\sec ^{2} x} \sec ^{2} x \cdot d x$

$I=\int_{0}^{\frac{\pi}{4}} \log (\tan x+1) d x$

$\bold{\left[\because \int_{0}^{a} f(x) d x=\int_{0}^{a} f(x-a) d x\right]}$

$I=\int_{0}^{\frac{\pi}{4}} \log \left(\tan \left(\frac{\pi}{4}-x\right)+1\right) d x$

$\left[\because \tan \left(\frac{\pi}{4}-x\right)=\frac{\tan \frac{\pi}{4}-\tan x}{1+\tan \frac{\pi}{4} \tan x}=\frac{1-\tan x}{1+\tan x}\right]$

$I=\int_{0}^{\frac{\pi}{4}} \log \left(\frac{1-\tan x}{1+\tan x}+1\right) d x$

$I=\int_{0}^{\frac{\pi}{4}} \log \left(\frac{1-\tan x+1+\tan x}{1+\tan x}\right) d x$

$I=\int_{0}^{\frac{\pi}{4}} \log \left(\frac{2}{1+\tan x}\right) d x$

$\bold{\left[\because \log \left(\frac{a}{b}\right)=\log a-\log b\right]}$

$I=\int_{0}^{\frac{\pi}{4}}(\log 2-\log (1+\tan x)) d x$

On substituting the limits, we get,

$I=\frac{\pi}{4} \log 2-I$

$2 I=\frac{\pi}{4} \log 2$

$\therefore I=\frac{\pi}{8} \log 2$