Definite Integration Ques 107
- For $x>0$, let $f(x)=\int _1^{x} \frac{\ln t}{1+t} d t$. Find the function $f(x)+f(1 / x)$ and show that $f(e)+f(1 / e)=1 / 2$, where $\ln t=\log _e t$.
$(2000,5 M)$
Show Answer
Answer:
Correct Answer: 107.$\frac{1}{2}(\ln x)^{2}$
Solution:
- $f(x)=\int _1^{x} \frac{\ln t}{1+t} d t$ for $x>0$
[given]
Now, $\quad f(1 / x)=\int _1^{1 / x} \frac{\ln t}{1+t} d t$
Put $t=1 / u \Rightarrow d t=\left(-1 / u^{2}\right) d u$
$$ \begin{aligned} \therefore \quad f(1 / x) & =\int _1^{x} \frac{\ln (1 / u)}{1+1 / u} \cdot \frac{(-1)}{u^{2}} d u \\ & =\int _1^{x} \frac{\ln u}{u(u+1)} d u=\int _1^{x} \frac{\ln t}{t(1+t)} d t \end{aligned} $$
Now, $f(x)+f \frac{1}{x}=\int _1^{x} \frac{\ln t}{(1+t)} d t+\int _1^{x} \frac{\ln t}{t(1+t)} d t$
$$ =\int _1^{x} \frac{(1+t) \ln t}{t(1+t)} d t=\int _1^{x} \frac{\ln t}{t} d t=\frac{1}{2}\left[(\ln t)^{2}\right] _1^{x}=\frac{1}{2}(\ln x)^{2} $$
Put $x=e$,
$$ f(e)+f \frac{1}{e}=\frac{1}{2}(\ln e)^{2}=\frac{1}{2} $$
Hence proved.
BETA
