Animated Solution for Mathematics - Definite Integration: Let $\alpha \in (0,1)$ and $\beta ={\log }_e(1-\alpha )$. Let $P_n(x)=x+\frac{x^2}{2}+\frac{x^3}{3}+...+\frac{x^n}{n},x\in (0,1)$. Then the integral ${\int }_0^{\alpha }\frac{t^{50}}{1-t}dt$ is equal to

Welcome, future engineers. Today, we are not just solving a calculus problem; we are embarking on a journey of mathematical elegance. In the JEE Advanced arena, you will often encounter problems that look like intimidating, monolithic expressions.

The secret to conquering them is not brute force, but rather the art of 'Algebraic Surgery'—the ability to dissect an expression, rearrange its components, and reveal the simple, beautiful structure hidden beneath.

Let us look at our target integral:

The Algebraic Surgery

At first glance, this looks like a nightmare. We have a high-power polynomial in the numerator and a linear term in the denominator. If you try to use standard integration by parts, you will be stuck in an infinite loop of complexity.

But pause. Take a breath. Look at the denominator: $1-t$.

In mathematics, whenever you see a denominator like $1-t$ or $1-x$, your intuition should immediately scream 'Geometric Series!' But to get there, we need to manipulate the numerator. We need to force a cancellation.

We perform surgery: we subtract $1$ and add $1$ to the numerator.

This simple act of adding and subtracting one is the turning point of the problem. It transforms our integral into:

By splitting this into two distinct integrals, we have effectively divided and conquered:

The Geometric Series Revelation

Now, let us focus on the first integral. The numerator is $t^{50}-1$. This is almost the geometric series form, but the signs are flipped. Let us factor out a negative sign:

Now, recall the identity for the sum of a geometric progression:

If we set $r=t$ and $n=50$, the expression becomes a beautiful, finite polynomial:

Suddenly, the 'nightmare' integral has vanished, replaced by a simple polynomial. Our integral $I$ now looks like this:

The Integration and the Logarithmic Trap

Integrating the polynomial part is straightforward. We apply the power rule, $\int t^{n}dt=\frac{t^{n+1}}{n+1}$, to each term. The result is:

When we evaluate this from $0$ to $\alpha$, the lower limit $0$ vanishes. We are left with exactly the definition of our polynomial $P_{50}(\alpha )$ given in the problem statement, but with a negative sign:

Now, we turn to the second integral: ${\int }_0^{\alpha }\frac{1}{1-t}dt$. Here is where many students stumble. Remember, the integral of $\frac{1}{1-t}$ is not just $\ln (1-t)$.

Because of the negative sign attached to $t$, we must account for the chain rule, which introduces another negative sign. The integral is $-\ln (1-t)$.

Evaluating this from $0$ to $\alpha$:

Since the problem defines $\beta ={\log }_e(1-\alpha )$, our result for this part is simply $-\beta$.

We have arrived at the finish line. We have two pieces of the puzzle: the polynomial part $-P_{50}(\alpha )$ and the logarithmic part $-\beta$.

Combining them, we get:

Factoring out the negative sign, we reach the elegant conclusion:

This matches our second option perfectly. Remember, students: the complexity of a problem is often just a mask. By using algebraic manipulation, recognizing series patterns, and being meticulous with signs, you can peel back that mask and find the simple truth underneath.