Animated Solution for Mathematics - Definite Integration: A value of $\alpha$ such that ${\int }_{\alpha }^{\alpha +1}\frac{dx}{(x+\alpha )(x+\alpha +1)}={\log }_e(9/8)$ is :

A Journey Through Integration

Welcome, future engineer! Today, we are going to peel back the layers of a seemingly intimidating integral.

When you first look at the problem ${\int }_{\alpha }^{\alpha +1}\frac{dx}{(x+\alpha )(x+\alpha +1)}=\ln (\frac{9}{8})$, it might look like a complex mess of variables. But I want you to take a deep breath.

In JEE Advanced, the most complex-looking problems often hide the most elegant, simple solutions. Let's embark on this journey together.

The Art of Deconstruction

The first thing that should catch your eye is the denominator: $(x+\alpha )(x+\alpha +1)$. Whenever you see a rational function with distinct linear factors in the denominator, your mind should immediately jump to partial fraction decomposition.

It is your secret weapon. Notice the difference between the two factors: $(x+\alpha +1)-(x+\alpha )=1$.

Because the numerator is also $1$, we can rewrite the integrand as:

See how the complexity just vanished? We have transformed a product into a simple difference. This is the power of mathematical intuition.

The Integration Journey

Now, let's substitute this back into our integral. We are now looking at:

Integrating this is straightforward. We know that $\int \frac{1}{x+k}dx=\ln |x+k|$.

Applying this to our terms, we get:

Before we plug in the limits, let's use the logarithmic property $\ln (a)-\ln (b)=\ln (\frac{a}{b})$ to make our lives easier. This gives us:

The Algebraic Climax

Now, we substitute the upper limit $x=\alpha +1$ and the lower limit $x=\alpha$.

For the upper limit:

For the lower limit:

Subtracting these gives us:

By equating the arguments, we get the quadratic equation:

Cross-multiplying leads us to $8(4{\alpha }^2+4\alpha +1)=9(4{\alpha }^2+4\alpha )$, which simplifies to $32{\alpha }^2+32\alpha +8=36{\alpha }^2+36\alpha$.

Rearranging this, we get $4{\alpha }^2+4\alpha -8=0$, or simply ${\alpha }^2+\alpha -2=0$. Factoring this gives $(\alpha +2)(\alpha -1)=0$.

Thus, $\alpha =-2$ or $\alpha =1$. Looking at our options, $\alpha =-2$ is the correct choice.

You see? The problem wasn't about brute-force integration; it was about recognizing the structure and simplifying it.

Keep practicing these patterns, and you will find that even the toughest JEE problems start to look like puzzles waiting to be solved. You've got this!