Definite Integration Question 18
Question 18
- 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 $\ln t=\log _{e} t$.
$(2000,5 \mathrm{M})$
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Answer:
Correct Answer: 18. $\frac{1}{2}(\ln x)^{2}$
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.
