Animated Solution for Mathematics - Indefinite Integration: Find integration $\int \frac{dx}{(x-3{)}^{6/7}(x+4{)}^{8/7}}$

Welcome, fellow traveler of the JEE Advanced path. Today, we are going to dismantle a problem that, at first glance, looks like a nightmare of radicals. We are looking at the integral:

When you see fractional powers like $6/7$ and $8/7$, your instinct might be to panic. But I want you to pause. In mathematics, complexity is often just a mask for a deeper, simpler truth.

Look at those exponents: $6/7$ and $8/7$. What happens when we add them?

This is not a coincidence; it is the heartbeat of the problem. Whenever you encounter an integral of the form $\int \frac{dx}{(x-a{)}^m(x-b{)}^n}$ where $m+n=2$, you are holding a golden key. The sum being an integer is your signal to perform a specific, elegant substitution.

We need to create a term that looks like a ratio, specifically $\frac{x-3}{x+4}$. To do this, we must manipulate the denominator. We have $(x-3{)}^{6/7}$. If we divide this by $(x+4{)}^{6/7}$, we get our desired ratio.

To maintain the balance of the equation, we multiply and divide the denominator by $(x+4{)}^{6/7}$:

Now, watch the magic happen. The product $(x+4{)}^{6/7}\cdot (x+4{)}^{8/7}$ combines to become $(x+4{)}^{14/7}$, which is $(x+4{)}^2$. Our integral is now:

Now that we have isolated the term ${\left(\frac{x-3}{x+4}\right)}^{6/7}$, let us define our new variable:

We need to find $dx$ in terms of $dt$. Using the quotient rule, we differentiate $t$ with respect to $x$:

This is the moment of triumph! We see that $\frac{dx}{(x+4{)}^2}=\frac{1}{7}dt$. The entire structure of the integral collapses into something incredibly simple.

We substitute everything back into the integral:

Applying the power rule, we get:

Finally, we substitute back $t=\frac{x-3}{x+4}$. Our final answer is:

See? The complexity vanished. You did not just solve a problem; you uncovered the underlying order of the math. Keep this technique in your arsenal, and no integral will ever intimidate you again.