Animated Solution for Mathematics - Definite Integration: Let $F(x)=f(x)+f(1/x)$, where $f(x)={\int }_1^x\frac{\log t}{1+t}dt$. Then $F(e)$ equals

Imagine you are standing before a mountain. The path straight up the middle looks steep, rocky, and impossible. That is exactly what this problem feels like at first glance.

We are given $F(x)=f(x)+f(1/x)$, where:

If you try to evaluate that integral directly, you will find yourself lost in a labyrinth of non-elementary functions. But here is the secret: in JEE Advanced, when a problem looks impossible to integrate, it is usually begging you to differentiate it instead.

We start by invoking the Newton-Leibniz rule. This is your best friend in calculus. It tells us that if we have a function defined by an integral with a variable upper limit, its derivative is simply the integrand evaluated at that limit.

For $f(x)={\int }_1^x\frac{\log t}{1+t}dt$, the derivative is immediate:

It is clean, elegant, and powerful. We have stripped away the integral sign in one stroke.

Now, we turn our attention to $F(x)=f(x)+f(1/x)$. We differentiate both sides with respect to $x$.

The first term is easy, but the second term, $f(1/x)$, requires the Chain Rule. We differentiate the outer function $f$ to get $f^{\prime }(1/x)$, and then multiply by the derivative of the inner function $1/x$, which is $-\frac{1}{x^2}$.

So, our derivative becomes:

Do not rush this step! That negative sign is where most students stumble.

Now, let us substitute our known $f^{\prime }(x)$ into the equation. We need $f^{\prime }(1/x)$, which is:

Using the property of logarithms, $\log (1/x)=-\log x$, and simplifying the denominator, we get:

When we plug this back into our expression for $F^{\prime }(x)$, the magic happens. The negative sign from the Chain Rule and the negative sign from the logarithm cancel out perfectly!

We are left with:

Factoring out $\frac{\log x}{1+x}$, we get $\frac{\log x}{1+x}\left(1+\frac{1}{x}\right)$, which simplifies beautifully to:

We have arrived at $F^{\prime }(x)=\frac{\log x}{x}$. This is a standard integral! By substituting $u=\log x$, we see that $du=\frac{1}{x}dx$.

The integral becomes:

Thus, $F(x)=\frac{1}{2}(\log x{)}^2+C$. To find $C$, we use the boundary condition $x=1$.

Since $f(1)={\int }_1^1\cdots =0$, we know $F(1)=f(1)+f(1)=0$. Plugging this in, we find $C=0$.

Finally, evaluating at $x=e$, we get:

You have conquered the mountain by finding the hidden path around it. Well done!