If f(x) = intlimits_1^x frac{ln t}{1 + t} dt | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(D) Given $f(x) = \int\limits_1^x \frac{\ln t}{1 + t} dt$.
Consider $f(1/x) = \int\limits_1^{1/x} \frac{\ln t}{1 + t} dt$.
Let $t = 1/u$,then $dt =

Vedclass · Accepted Answer

All $x > 0$

Vedclass · Answer

(D) Given $f(x) = \int\limits_1^x \frac{\ln t}{1 + t} dt$.
Consider $f(1/x) = \int\limits_1^{1/x} \frac{\ln t}{1 + t} dt$.
Let $t = 1/u$,then $dt = -1/u^2 du$.
When $t = 1$,$u = 1$,and when $t = 1/x$,$u = x$.
Substituting these into the integral:
$f(1/x) = \int\limits_1^x \frac{\ln(1/u)}{1 + 1/u} (-1/u^2) du = \int\limits_1^x \frac{-\ln u}{\frac{u+1}{u}} \cdot \frac{1}{u^2} du = -\int\limits_1^x \frac{\ln u}{u(u+1)} du$.
Now,$f(x) + f(1/x) = \int\limits_1^x \frac{\ln t}{1 + t} dt - \int\limits_1^x \frac{\ln t}{t(1 + t)} dt = \int\limits_1^x \ln t \left( \frac{1}{1+t} - \frac{1}{t(1+t)} \right) dt$.
Simplifying the integrand:
$\frac{1}{1+t} - \frac{1}{t(1+t)} = \frac{t-1}{t(1+t)}$.
Thus,$f(x) + f(1/x) = \int\limits_1^x \frac{\ln t (t-1)}{t(1+t)} dt$.
For this integral to be $0$ for all $x$,the integrand must be $0$,which is not true. However,if the question implies the identity $f(x) + f(1/x) = \frac{1}{2}(\ln x)^2$,then the equation $f(x) + f(1/x) = 0$ is only satisfied at $x = 1$. Given the options,if the equation is an identity,it is not satisfied for any specific $x$ other than $1$. If the question is interpreted as $f(x) + f(1/x) = 0$,then $x=1$ is the only solution. Since $x=1$ is not explicitly listed,the correct choice is $(D)$.

If $f(x) = \int\limits_1^x \frac{\ln t}{1 + t} dt$ where $x > 0$,then the value$(s)$ of $x$ satisfying the equation $f(x) + f(1/x) = 0$ is/are:

Similar Questions

Let $g(x) = \int_{0}^{x} f(t) dt$,where $f$ is a continuous function in $[0, 3]$ such that $\frac{1}{3} \leq f(t) \leq 1$ for all $t \in [0, 1]$ and $0 \leq f(t) \leq \frac{1}{2}$ for all $t \in (1, 3]$. The largest possible interval in which $g(3)$ lies is:

$\int_{0}^{1} x(1-x)^{5} dx =$

$\int_0^{1000} e^{x-[x]} \, dx$ is equal to

If the integral $\int_{0}^{10} \frac{[\sin 2 \pi x ]}{ e ^{ x -[ x ]}} dx =\alpha e ^{-1}+\beta e ^{-\frac{1}{2}}+\gamma$,where $\alpha, \beta, \gamma$ are integers and $[ x ]$ denotes the greatest integer less than or equal to $x$,then the value of $\alpha+\beta+\gamma$ is equal to ........ .

$\int_0^\pi \frac{x \tan x}{\sec x+\tan x} d x$ is equal to

Vedclass Test Series

Exam Paper Generator

Online Exam Module