Exam $9768$ of subject $ GRE Q1.$

Question

This is the question:

Could anyone give me a hint on how can I solve it?

Answer 1

Hint: Use the Leibniz´s rule to compute the derivative $$\frac{d}{dx} \int_{u(x)}^{v(x)}f(x,t) \, dt =f(x,v(x))\cdot \frac{dv}{dx}-f(x,u(x))\cdot \frac{du}{dx}+\int_{_{}^{u(x)}}^{v(x)} \frac{\partial f(x,t)}{\partial x} \, dt$$

with $u(x)=e, v(x)=x, f(x,t)=\log(t)$

In your case $f(x,t)$ does not depend on $x$. Thus $\frac{\partial f(x,t)}{\partial x}=0$. And $\frac{du}{dx}=0$ as well.

Answer 2

HINT :

let $G$ be a primitive of $g(t) = log(t)$ then $F(x) = G(x)-G(e)$

What is $F'$ ? Can you finish ?

Answer 3

Use the Leibniz Integral Rule
$$\frac{d}{dx}\int_{g(x)}^{h(x)}f(t)dt=f(h(x)).\frac{dh}{dx}-f(g(x)).\frac{dg}{dx}$$
Given $f(t)=\ln(t)$, $h(x)=x$, $g(x)=e$

Answer 4

The integral $\int \log x\ dx$ is just $x\log x-x$. If you apply the upper and lower bounds in your case, the result is $(x \log x -x - e\log e +e)$ (with $e$ being the Euler's constant).

Now you just have to differentiate this w.r.t. $x$. This is easy and in the end you get $\log x$.

This is a step by step method required that you know the primitive of $\log x$.