If ∫ 1 / ⁵√((x-1)⁴(x+3)⁶) dx = A ((αx - 1)/(βx + 3))ᴮ + C, where C is the constant of integration, t

Q: Wait, why did we multiply and divide by $(x+3)^{4/5}$? How would I ever think of that in an exam?

That is a fantastic question! In JEE Advanced, we often encounter integrals where the denominator is a product of powers of linear factors. The trick is to look at the sum of the exponents. If the sum is an integer (like 2 here), you can force the expression into a form where you have a ratio raised to a power multiplied by a polynomial. It is a standard technique for 'algebraic manipulation before calculus.' Practice recognizing these exponent patterns, and it will become second nature.

Q: Why is the derivative of $\frac{x-1}{x+3}$ so convenient? Is it always this clean?

It is not always this clean, but it is designed to be! When you have a linear fractional transformation of the form $\frac{ax+b}{cx+d}$, the derivative will always have a denominator of $(cx+d)^2$. If your integral contains a $(cx+d)^2$ in the denominator, this substitution is almost certainly the intended path. It is a beautiful piece of mathematical symmetry.

Q: Can I use a different substitution, like $u = x-1$? It seems more direct.

You could try, but you will quickly find yourself in a swamp of algebra. Substituting $u = x-1$ would leave you with terms like $(u+4)^{6/5}$ in the denominator, which does not simplify the integral. The goal of a good substitution is to collapse the complexity, not just change the variable. Always look for the substitution that matches the structure of the target form given in the problem.

Question

If $\int \frac{1}{\sqrt[5]{(x-1)^4(x+3)^6}} dx = A \left( \frac{\alpha x - 1}{\beta x + 3} \right)^B + C$, where $C$ is the constant of integration, then the value of $\alpha + \beta + 20AB$ is __________

Sigma Percentile · Accepted Answer

### Analyzing the Setup

Welcome, fellow traveler on the path to JEE mastery. Today, we are going to dismantle a problem that looks like a nightmare of radicals and fractions but is, in reality, a masterclass in algebraic elegance.

We are tasked with evaluating the integral:

$$I = \int \frac{1}{\sqrt[5]{(x-1)^4(x+3)^6}} dx$$

We must match this to the form $A \left( \frac{\alpha x - 1}{\beta x + 3} \right)^B + C$. When you see a problem like this, your first instinct might be panic, but take a deep breath. In the world of JEE Advanced, complexity is often just a mask for a hidden, simpler structure.

### The Art of Manipulation

Let us start by stripping away the radical. We know that $\sqrt[5]{(x-1)^4(x+3)^6}$ is equivalent to $(x-1)^{4/5}(x+3)^{6/5}$.

Our integral becomes:

$$I = \int \frac{1}{(x-1)^{4/5}(x+3)^{6/5}} dx$$

Now, observe the exponents: $4/5$ and $6/5$. Their sum is $4/5 + 6/5 = 10/5 = 2$. This is a breadcrumb left by the examiner to help us create a perfect square.

By multiplying and dividing the denominator by $(x+3)^{4/5}$, we obtain:

$$(x-1)^{4/5}(x+3)^{6/5} = \frac{(x-1)^{4/5}}{(x+3)^{4/5}} \cdot (x+3)^{6/5} \cdot (x+3)^{4/5}$$

This simplifies beautifully to:

$$\left( \frac{x-1}{x+3} \right)^{4/5} (x+3)^2$$

### The Elegant Substitution

Now that we have the integral in the form:

$$I = \int \frac{1}{\left( \frac{x-1}{x+3} \right)^{4/5} (x+3)^2} dx$$

The path forward is illuminated. We define our substitution as $t = \frac{x-1}{x+3}$.

Using the quotient rule to differentiate $t$ with respect to $x$:

$$\frac{dt}{dx} = \frac{(x+3)(1) - (x-1)(1)}{(x+3)^2} = \frac{x+3-x+1}{(x+3)^2} = \frac{4}{(x+3)^2}$$

This confirms that $\frac{1}{(x+3)^2} dx = \frac{1}{4} dt$. Everything fits perfectly.

### The Collapse and the Victory

With our substitution ready, the integral collapses into a standard power rule problem:

$$I = \int \frac{1}{t^{4/5}} \left( \frac{1}{4} dt \right) = \frac{1}{4} \int t^{-4/5} dt$$

Integrating $t^{-4/5}$ gives us $\frac{t^{1/5}}{1/5} = 5t^{1/5}$. Multiplying by the constant $1/4$, we arrive at:

$$I = \frac{5}{4} t^{1/5} + C$$

Substituting back $t = \frac{x-1}{x+3}$, we get:

$$I = \frac{5}{4} \left( \frac{x-1}{x+3} \right)^{1/5} + C$$

### The Final Comparison

We compare our result $\frac{5}{4} \left( \frac{1 \cdot x - 1}{1 \cdot x + 3} \right)^{1/5} + C$ with the target form $A \left( \frac{\alpha x - 1}{\beta x + 3} \right)^B + C$.

By direct inspection, we identify:
$A = 5/4$, $B = 1/5$, $\alpha = 1$, and $\beta = 1$.

The final step is to calculate the value of $\alpha + \beta + 20AB$:

$$1 + 1 + 20 \cdot \left( \frac{5}{4} \right) \cdot \left( \frac{1}{5} \right) = 2 + 5 = 7$$

The final result is **7**. A problem that seemed daunting at the start has been reduced to a sequence of logical, elegant steps.

Animated Solution for Mathematics - Indefinite Integration: If $\int \frac{1}{5 ( x - 1 ) ^{4} ( x + 3 ) ^{6}} d x = A (\frac{α x - 1}{β x + 3})^{B} + C$ , where $C$ is the constant of integration, then the value of $α + β + 20 A B$ is __________

Enter Numerical Value:

Visualized Solution

Analyze the Integral Form

Rewrite the Integrand

Manipulate the Denominator

Simplified Integral Form

Define Substitution $t$

Differentiate the Substitution

Express $d x$ in terms of $d t$

Substitute into the Integral

Apply Integration Power Rule

Simplify the Coefficient

Back-Substitution

Compare with Given Form

Calculate Final Value

The Sigma Insight: Integration by Substitution

Analyzing the Setup

The Art of Manipulation

The Elegant Substitution

The Collapse and the Victory

The Final Comparison

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