The integral ∫ 1 / ⁴√((x-1)³(x+2)⁵) dx is equal to : (where C is a constant of integration)
(1) 3/4

Q: Why did we choose to multiply by $(x-1)^{5/4}$ specifically? Why not something else?

Great question! We chose $(x-1)^{5/4}$ because we wanted the exponent of $(x-1)$ to match the exponent of $(x+2)$, which is $5/4$. This allows us to combine them into a single ratio $\left(\frac{x+2}{x-1}\right)^{5/4}$. If we chose a different power, we wouldn't be able to group them cleanly.

Q: Wait, why did the derivative of $\frac{x+2}{x-1}$ result in a negative sign? Did I mess up the quotient rule?

It is a common point of confusion! Remember the quotient rule: $\frac{d}{dx} \left(\frac{u}{v}\right) = \frac{v u' - u v'}{v^2}$. Here, $u = x+2$ and $v = x-1$. So, $v u' - u v' = (x-1)(1) - (x+2)(1) = x - 1 - x - 2 = -3$. The negative sign is correct and essential for the final answer.

Q: What if the sum of the exponents was not an integer? Does this method still work?

This specific trick relies on the sum of the exponents being an integer. If the sum is not an integer, the substitution $t = \frac{x+2}{x-1}$ won't simplify the integral into a standard power rule form. In such cases, you might need different techniques like trigonometric substitution or other specialized forms. Always check the exponent sum first!

Question

The integral $\int \frac{1}{\sqrt[4]{(x-1)^3(x+2)^5}} dx$ is equal to : (where $C$ is a constant of integration)

Options:

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

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

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

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

Sigma Percentile · Accepted Answer

### Analyzing the Setup

The integral $I = \int \frac{1}{\sqrt[4]{(x-1)^3(x+2)^5}} dx$ appears daunting due to the presence of radicals and fractional powers. However, in JEE Advanced mathematics, such structures often hide a specific algebraic symmetry.

We begin by rewriting the radical expression using fractional exponents:

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

Observe the sum of the exponents: $\frac{3}{4} + \frac{5}{4} = \frac{8}{4} = 2$. Since this sum is an integer, it serves as a clear indicator that we can simplify the expression by introducing a ratio of the two linear factors.

### The Algebraic Surgery

To force the appearance of the ratio $\frac{x+2}{x-1}$, we multiply and divide the denominator by $(x-1)^{5/4}$. This allows us to group the terms effectively:

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

Since $(x-1)^{3/4} \cdot (x-1)^{5/4} = (x-1)^2$, the integral simplifies significantly:

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

### The Transformation

We now define our substitution variable as $t = \frac{x+2}{x-1}$. To find the differential $dt$, we apply the quotient rule:

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

This yields the crucial relation $\frac{dx}{(x-1)^2} = -\frac{1}{3} dt$. Substituting these into our integral, the expression collapses into a standard power form:

$$I = \int \frac{1}{t^{5/4}} \left( -\frac{1}{3} dt \right) = -\frac{1}{3} \int t^{-5/4} dt$$

### Final Calculation

Applying the power rule for integration, we obtain:

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

Substituting $t = \frac{x+2}{x-1}$ back into the equation, we get:

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

By flipping the fraction to eliminate the negative exponent, we arrive at the **final answer**:

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

Animated Solution for Mathematics - Indefinite Integration: The integral $\int \frac{1}{4 ( x - 1 ) ^{3} ( x + 2 ) ^{5}} d x$ is equal to : (where $C$ is a constant of integration)

Select Answer:

Visualized Solution

Analyze the Integral Structure

Strategy: Creating a Ratio

Simplify the Denominator

Define the Substitution

Differentiate to find $d t$

Substitute into the Integral

Integrate with respect to $t$

Final Back-substitution

Conclusion and Key Takeaway

The Sigma Insight: Integration by Substitution

Analyzing the Setup

The Algebraic Surgery

The Transformation

Final Calculation

Similar Questions

The integral $\int \frac{d x}{x ^{2} ( x ^{4} + 1 ) ^{3/4}}$ equals

$\int \frac{x ^{2} - 1}{x ^{3} 2 x ^{4} - 2 x ^{2} + 1} d x =$

The integral $\int \frac{2 x ^{3} - 1}{x ^{4} + x} d x$ is equal to :

The integral $\int \frac{2 x ^{12} + 5 x ^{9}}{( x ^{5} + x ^{3} + 1 ) ^{3}} d x$ is equal to

If $\int \frac{x + 1}{2 x - 1} d x = f (x) 2 x - 1 + C$ , where $C$ is a constant of integration, then $f (x)$ is equal to:

If $f (\frac{x - 4}{x + 2}) = 2 x + 1, (x \in R - {1, - 2})$ , then $\int f (x) d x$ is equal to : (where C is a constant of integration)

The integral $\int \frac{d x}{( x + 4 ) ^{8/7} ( x - 3 ) ^{6/7}}$ is equal to: (where C is a constant of integration)

Evaluate the following $\int \frac{d x}{x ^{2} ( x ^{4} + 1 ) ^{3/4}}$

Find the indefinite integral $\int (\frac{1}{3 x + 4 x} + \frac{l n ( 1 + 6 x )}{3 x + x}) d x$

For $x^{2} \neq = nπ + 1$ , $n \in N$ (the set of natural numbers), the integral $\int \frac{2 s i n ( x ^{2} - 1 ) - s i n 2 ( x ^{2} - 1 )}{2 s i n ( x ^{2} - 1 ) + s i n 2 ( x ^{2} - 1 )} d x$ is equal to : (where c is a constant of integration)

Animated Solution for Mathematics - Indefinite Integration: .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } The integral ∫4(x−1)3(x+2)5​1​dx is equal to : (where C is a constant of integration)