SATHEE: Indefinite Integration 3 Question 18

Indefinite Integration 3 Question 18

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 $\ln t = \log_{e} t$ .

$(2000, 5 M)$

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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, $f (1 / x) = \int_{1}^{1 / x} \frac{\ln t}{1 + t} d t$

Put $t = 1 / u \Rightarrow d t = (- 1 / u^{2}) d u$

$\begin{aligned} ∴ f (1 / x) & = \int_{1}^{x} \frac{\ln (1 / u)}{1 + 1 / u} \cdot \frac{(- 1)}{u^{2}} d u \\ = \int_{1}^{x} \frac{\ln u}{u (u + 1)} d u = \int_{1}^{x} \frac{\ln t}{t (1 + t)} d t \end{aligned}$

Now, $f (x) + f \frac{1}{x} = \int_{1}^{x} \frac{\ln t}{(1 + t)} d t + \int_{1}^{x} \frac{\ln t}{t (1 + t)} d t$

$= \int_{1}^{x} \frac{(1 + t) \ln t}{t (1 + t)} d t = \int_{1}^{x} \frac{\ln t}{t} d t = \frac{1}{2} {[(\ln t)^{2}]}_{1}^{x} = \frac{1}{2} (\ln x)^{2}$

Put $x = e$ ,

$f (e) + f \frac{1}{e} = \frac{1}{2} (\ln e)^{2} = \frac{1}{2}$

Hence proved.

Indefinite Integration 3 Question 18

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 $\ln t = \log_{e} t$ .

Answer:

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Indefinite Integration 3 Question 18

18. For x>0, let f(x)=∫1xln⁡t1+tdt. Find the function f(x)+f(1/x) and show that f(e)+f(1/e)=1/2, where ln⁡t=loge⁡t.

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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 $\ln t = \log_{e} t$ .