JEE Main 2025 (January)

LEVELJEE Main

Animated Solution for Mathematics - Indefinite Integration: Let $I (x) = \int \frac{d x}{( x - 11 ) ^{\frac{1}{13}} ( x + 15 ) ^{\frac{15}{13}}} . I f I (37) - I (24) = \frac{1}{4} (\frac{1}{b ^{\frac{1}{13}}} - \frac{1}{c ^{\frac{1}{13}}})$ , b, $c \in N,$ then $3 (b + c)$ is equal to

Q: Why do we get $1/4$ as the coefficient?

After substituting $t = \frac{x-11}{x+15}$, the integral becomes $\frac{1}{26} \int t^{-11/13} dt$. Integrating $t^{-11/13}$ gives $\frac{t^{2/13}}{2/13}$. Multiplying this by the constant $1/26$ gives $\frac{1}{26} \cdot \frac{13}{2} = \frac{1}{4}$. It is a beautiful simplification where the constants align perfectly.

Select Answer:

Visualized Solution

Analyze the Integrand

Given integral: $I (x) = \int \frac{d x}{( x - 11 ) ^{\frac{11}{13}} ( x + 15 ) ^{\frac{15}{13}}}$
Notice the fractional powers in the denominator.

The Power Sum Trick

Observe the sum of the powers in the denominator: $\frac{11}{13} + \frac{15}{13}$
Sum $= \frac{26}{13} = 2$
This integer sum is a massive hint for a specific algebraic manipulation.

Creating the Ratio

We need to create the term $\frac{x - 11}{x + 15}$ in the denominator.
Multiply and divide the denominator by $(x + 15)^{\frac{11}{13}}$ :
$I (x) = \int \frac{d x}{( \frac{x - 11}{x + 15} ) ^{\frac{11}{13}} ( x + 15 ) ^{\frac{11}{13}} ( x + 15 ) ^{\frac{15}{13}}}$

Simplifying the Denominator

Combine the powers of $(x + 15)$ :
$\frac{11}{13} + \frac{15}{13} = 2$
The integral becomes: $I (x) = \int \frac{d x}{( \frac{x - 11}{x + 15} ) ^{\frac{11}{13}} ( x + 15 ) ^{2}}$

Applying Substitution

Let $t = \frac{x - 11}{x + 15}$
This substitution will simplify the complex fractional power.

Differentiating the Substitution

Differentiate $t$ with respect to $x$ using the quotient rule:
$\frac{d t}{d x} = \frac{1 \cdot ( x + 15 ) - 1 \cdot ( x - 11 )}{( x + 15 ) ^{2}}$
$\frac{d t}{d x} = \frac{x + 15 - x + 11}{( x + 15 ) ^{2}} = \frac{26}{( x + 15 ) ^{2}}$
Therefore, $\frac{d x}{( x + 15 ) ^{2}} = \frac{d t}{26}$

Transforming the Integral

Substitute $t$ and $d t$ into the integral:
$I (x) = \int \frac{1}{t ^{\frac{11}{13}}} \cdot \frac{d t}{26}$
$I (x) = \frac{1}{26} \int t^{- \frac{11}{13}} d t$

Performing the Integration

Apply the power rule: $\int t^{n} d t = \frac{t ^{n + 1}}{n + 1}$
$I (x) = \frac{1}{26} (\frac{t ^{- \frac{11}{13} + 1}}{- \frac{11}{13} + 1}) + C$
$I (x) = \frac{1}{26} (\frac{t ^{\frac{2}{13}}}{\frac{2}{13}}) + C$
$I (x) = \frac{1}{26} \cdot \frac{13}{2} t^{\frac{2}{13}} + C = \frac{1}{4} t^{\frac{2}{13}} + C$

Back Substitution

Substitute $t = \frac{x - 11}{x + 15}$ back into the expression:
$I (x) = \frac{1}{4} (\frac{x - 11}{x + 15})^{\frac{2}{13}} + C$

Evaluating $I (37)$

Calculate $I (37)$ :
$I (37) = \frac{1}{4} (\frac{37 - 11}{37 + 15})^{\frac{2}{13}} + C$
$I (37) = \frac{1}{4} (\frac{26}{52})^{\frac{2}{13}} + C = \frac{1}{4} (\frac{1}{2})^{\frac{2}{13}} + C$

Evaluating $I (24)$

Calculate $I (24)$ :
$I (24) = \frac{1}{4} (\frac{24 - 11}{24 + 15})^{\frac{2}{13}} + C$
$I (24) = \frac{1}{4} (\frac{13}{39})^{\frac{2}{13}} + C = \frac{1}{4} (\frac{1}{3})^{\frac{2}{13}} + C$

Finding the Difference

Find the difference $I (37) - I (24)$ :
$I (37) - I (24) = \frac{1}{4} [(\frac{1}{2})^{\frac{2}{13}} - (\frac{1}{3})^{\frac{2}{13}}]$
Rewrite to match the form $\frac{1}{4} (\frac{1}{b ^{\frac{1}{13}}} - \frac{1}{c ^{\frac{1}{13}}})$ :
$(\frac{1}{2})^{\frac{2}{13}} = \frac{1}{( 2 ^{2} ) ^{\frac{1}{13}}} = \frac{1}{4 ^{\frac{1}{13}}}$
$(\frac{1}{3})^{\frac{2}{13}} = \frac{1}{( 3 ^{2} ) ^{\frac{1}{13}}} = \frac{1}{9 ^{\frac{1}{13}}}$

Final Calculation

Comparing with $\frac{1}{4} (\frac{1}{b ^{\frac{1}{13}}} - \frac{1}{c ^{\frac{1}{13}}})$ gives $b = 4$ and $c = 9$ .
Calculate the final required value: $3 (b + c)$
$3 (b + c) = 3 (4 + 9) = 3 (13) = 39$
The correct option is (2).

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The Sigma Insight: Evaluation of Special Integral Forms

Analyzing the Setup

Imagine you are standing before an integral that looks like a tangled mess of fractional powers. You see:

I (x) = \int \frac{d x}{( x - 11 ) ^{11/13} ( x + 15 ) ^{15/13}}

It is intimidating, isn't it? But in the world of JEE Advanced, complexity is often just a mask for elegance.

The first step in our journey is to look for the hidden pattern. Notice the exponents:

11/13

and

15/13

. When you add them, you get

26/13

, which is exactly

2

This is not a coincidence; it is a signpost. Whenever you see this, it is a signal to create a rational function substitution.

The Art of Substitution

To unlock this, we need to force a ratio. We want to see

(\frac{x - 11}{x + 15})

appear.

We multiply the denominator by

(x + 15)^{11/13}

and, to keep the balance, we multiply the numerator by the same. This transforms our integral into:

I (x) = \int \frac{d x}{( \frac{x - 11}{x + 15} ) ^{11/13} ( x + 15 ) ^{2}}

Now, the magic happens. We set

t = \frac{x - 11}{x + 15}

Using the quotient rule, the derivative

\frac{d t}{d x}

becomes:

\frac{d t}{d x} = \frac{( x + 15 ) ( 1 ) - ( x - 11 ) ( 1 )}{( x + 15 ) ^{2}} = \frac{26}{( x + 15 ) ^{2}}

This means

\frac{d x}{( x + 15 ) ^{2}} = \frac{d t}{26}

. Our integral, which once seemed impossible, now simplifies to:

I (x) = \frac{1}{26} \int t^{- 11/13} d t

The Final Stretch

Applying the power rule, we integrate

t^{- 11/13}

to get

\frac{t ^{2/13}}{2/13}

. Multiplying by the constant

1/26

, we arrive at:

I (x) = \frac{1}{4} t^{2/13} + C

Substituting back

t = \frac{x - 11}{x + 15}

, we have our general solution. Evaluating

I (37)

and

I (24)

gives us:

I (37) = \frac{1}{4} (\frac{37 - 11}{37 + 15})^{2/13} = \frac{1}{4} (\frac{26}{52})^{2/13} = \frac{1}{4} (\frac{1}{2})^{2/13}

I (24) = \frac{1}{4} (\frac{24 - 11}{24 + 15})^{2/13} = \frac{1}{4} (\frac{13}{39})^{2/13} = \frac{1}{4} (\frac{1}{3})^{2/13}

The difference is

\frac{1}{4} [(\frac{1}{4})^{1/13} - (\frac{1}{9})^{1/13}]

. By comparing this to the given form, we find $b = 4$ and $c = 9$ .

The final calculation

3 (4 + 9)

leads us to $39$ . A truly elegant journey from chaos to order.

Common Traps & FAQs

Why does the sum of the powers being an integer matter?

In integration problems involving fractional powers, having the sum of exponents equal an integer is a classic sign that a substitution will rationalize the integrand. It allows us to group the terms into a single ratio, effectively removing the fractional complexity.

How did you know to multiply by

(x + 15)^{11/13}

We observed that the denominator had two terms with fractional powers. By multiplying by

(x + 15)^{11/13}

, we combined the powers of

(x + 15)

to get

(x + 15)^{11/13 + 15/13} = (x + 15)^{2}

. This created a perfect square in the denominator, which is exactly what we need for the derivative of our substitution

t = \frac{x - 11}{x + 15}

Why do we get

1/4

as the coefficient?

After substituting

t = \frac{x - 11}{x + 15}

, the integral becomes

\frac{1}{26} \int t^{- 11/13} d t

. Integrating

t^{- 11/13}

gives

\frac{t ^{2/13}}{2/13}

. Multiplying this by the constant

1/26

gives

\frac{1}{26} \cdot \frac{13}{2} = \frac{1}{4}

. It is a beautiful simplification where the constants align perfectly.

Animated Solution for Mathematics - Indefinite Integration: Let $I (x) = \int \frac{d x}{( x - 11 ) ^{\frac{1}{13}} ( x + 15 ) ^{\frac{15}{13}}} . I f I (37) - I (24) = \frac{1}{4} (\frac{1}{b ^{\frac{1}{13}}} - \frac{1}{c ^{\frac{1}{13}}})$ , b, $c \in N,$ then $3 (b + c)$ is equal to

Select Answer:

Visualized Solution

Analyze the Integrand

The Power Sum Trick

Creating the Ratio

Simplifying the Denominator

Applying Substitution

Differentiating the Substitution

Transforming the Integral

Performing the Integration

Back Substitution

Evaluating $I (37)$

Evaluating $I (24)$

Finding the Difference

Final Calculation

The Sigma Insight: Evaluation of Special Integral Forms

Analyzing the Setup

The Art of Substitution

The Final Stretch

Similar Questions

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Select Answer:

Visualized Solution

Analyze the Integrand

The Power Sum Trick

Creating the Ratio

Simplifying the Denominator

Applying Substitution

Differentiating the Substitution

Transforming the Integral

Performing the Integration

Back Substitution

Evaluating .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } I(37)

Evaluating .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } I(24)

Finding the Difference

Final Calculation

The Sigma Insight: Evaluation of Special Integral Forms

Analyzing the Setup

The Art of Substitution

The Final Stretch

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