JEE Main 2025 (January)

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Animated Solution for Mathematics - Indefinite Integration: If $f (x) = \int \frac{1}{x ^{1/4} ( 1 + x ^{1/4} )} d x$ , $f (0) = - 6,$ then $f (1)$ is equal to:

Q: Wait, why do we use $x^{1/4} = t$ instead of just $x = t^4$ directly?

That is a great question! While $x = t^4$ is the same substitution, setting $t = x^{1/4}$ is often more intuitive because it directly addresses the fractional power that makes the integral look complicated. By defining $t$ as the root, we simplify the expression $x^{1/4}$ to $t$ immediately, making the substitution process cleaner and less prone to algebraic errors.

Q: Why do we reject the negative root or absolute value in the final log term?

In this specific problem, we have $t = x^{1/4}$. Since $x$ is typically assumed to be non-negative in the context of real-valued functions with fractional powers like $x^{1/4}$, $t$ will also be non-negative. Therefore, $1+t$ is always positive, which allows us to drop the absolute value bars in $\ln|1+t|$ and write it simply as $\ln(1+t)$.

Select Answer:

Progressive Hints

Hint 1

Hint 1: The Spark. The presence of

x^{1/4}

in both the numerator and the denominator is a clear signal. Instead of dealing with fractional powers directly, visualize a substitution that transforms the entire integrand into a rational function. What if you set

t = x^{1/4}

? How does this change the differential

d x

Hint 2

Hint 2: The Arsenal. Once you have substituted

x = t^{4}

and

d x = 4 t^{3} d t

, you will arrive at an integral of the form

\int \frac{4 t ^{3}}{t ( 1 + t )} d t

. Simplify this by canceling the common factor of

t

. Now, you are left with

\int \frac{4 t ^{2}}{1 + t} d t

. To integrate this, use the algebraic trick of adding and subtracting 1 in the numerator:

t^{2} = (t^{2} - 1) + 1

Hint 3

Hint 3: The Execution. After splitting the integral, you will get

4 \int (t - 1 + \frac{1}{1 + t}) d t

. Integrate this term by term to get

4 (\frac{t ^{2}}{2} - t + ln ∣1 + t ∣) + C

. Now, substitute

t = x^{1/4}

back into the expression and use the condition

f (0) = - 6

to find the value of the constant

C

Visualized Solution

Substitution Strategy: $x^{\frac{1}{4}} = t$

Given function: $f (x) = \int \frac{1}{x ^{\frac{1}{4}} ( 1 + x ^{\frac{1}{4}} )} d x$
Let's use substitution: $x^{\frac{1}{4}} = t$
This implies: $x = t^{4}$

Calculating the Differential $d x$

Differentiate $x = t^{4}$ with respect to $t$
$\frac{d x}{d t} = 4 t^{3}$
So, $d x = 4 t^{3} d t$

Transforming the Integral

Substitute $d x$ and $x^{\frac{1}{4}}$ into the integral:
$I = \int \frac{4 t ^{3}}{t ( 1 + t )} d t$
Simplify by canceling $t$ : $I = \int \frac{4 t ^{2}}{1 + t} d t$

Algebraic Manipulation of the Integrand

Manipulate the numerator: $\frac{t ^{2}}{1 + t} = \frac{t ^{2} - 1 + 1}{1 + t}$
Factorize: $\frac{( t - 1 ) ( t + 1 ) + 1}{1 + t}$
Divide term by term: $(t - 1) + \frac{1}{1 + t}$

Integrating Term by Term

Integrate the expression: $4 \int (t - 1 + \frac{1}{1 + t}) d t$
Apply integration rules: $4 [\frac{t ^{2}}{2} - t + ln ∣1 + t ∣] + C$
Distribute the constant: $2 t^{2} - 4 t + 4 ln ∣1 + t ∣ + C$

Back-substitution to $x$

Substitute $t = x^{\frac{1}{4}}$ back into the equation
$t^{2} = (x^{\frac{1}{4}})^{2} = x^{\frac{1}{2}}$
$f (x) = 2 x^{\frac{1}{2}} - 4 x^{\frac{1}{4}} + 4 ln ∣1 + x^{\frac{1}{4}} ∣ + C$

Finding the Constant $C$

Use the given condition: $f (0) = - 6$
Substitute $x = 0$ : $2 (0)^{\frac{1}{2}} - 4 (0)^{\frac{1}{4}} + 4 ln ∣1 + 0∣ + C = - 6$
Simplify: $0 - 0 + 4 ln (1) + C = - 6$
Since $ln (1) = 0$ , we get $C = - 6$

Calculating $f (1)$

Substitute $x = 1$ and $C = - 6$ into $f (x)$
$f (1) = 2 (1)^{\frac{1}{2}} - 4 (1)^{\frac{1}{4}} + 4 ln ∣1 + 1∣ - 6$
Simplify the terms: $f (1) = 2 - 4 + 4 ln (2) - 6$

Final Result and Conclusion

Combine the constant terms: $2 - 4 - 6 = - 8$
$f (1) = 4 ln (2) - 8$
Factor out $4$ : $f (1) = 4 (ln (2) - 2)$
Final Answer matches option A: $4 (lo g_{e} 2 - 2)$

00:00 / 00:00

The Sigma Insight: Integration by Substitution

Analyzing the Setup

The integral we are tasked with solving is:

f (x) = \int \frac{1}{x ^{1/4} ( 1 + x ^{1/4} )} d x

When encountering fractional powers like

x^{1/4}

, the most effective strategy is to simplify the variable through substitution. This clears the path for standard algebraic manipulation.

The Key of Substitution

Let us set

x^{1/4} = t

. This implies

x = t^{4}

To find the differential

d x

, we differentiate

x = t^{4}

with respect to

t

\frac{d x}{d t} = 4 t^{3} ⟹ d x = 4 t^{3} d t

This substitution is the essential key; it removes the fractional exponents and prepares the expression for integration.

The Algebraic Dance

Substituting these values into our integral, we obtain:

I = \int \frac{4 t ^{3}}{t ( 1 + t )} d t

By canceling the common factor of

t

in the numerator and denominator, the expression simplifies to:

I = \int \frac{4 t ^{2}}{1 + t} d t

Since the degree of the numerator is higher than the denominator, we perform polynomial division or algebraic manipulation. We rewrite

t^{2}

(t^{2} - 1) + 1

I = 4 \int \frac{( t - 1 ) ( t + 1 ) + 1}{1 + t} d t

Splitting the fraction yields:

I = 4 \int (t - 1 + \frac{1}{1 + t}) d t

The Calculus Core

Integrating term by term, we get:

I = 4 (\frac{t ^{2}}{2} - t + ln ∣1 + t ∣) + C

Distributing the constant 4, we arrive at:

f (t) = 2 t^{2} - 4 t + 4 ln ∣1 + t ∣ + C

Returning to the original variable

x

by substituting

t = x^{1/4}

f (x) = 2 x^{1/2} - 4 x^{1/4} + 4 ln ∣1 + x^{1/4} ∣ + C

Final Calculation

We are given the initial condition

f (0) = - 6

. Substituting

x = 0

into our result:

2 (0)^{1/2} - 4 (0)^{1/4} + 4 ln ∣1 + 0∣ + C = - 6

Since

ln (1) = 0

, we find that

C = - 6

. Now, we evaluate

f (1)

f (1) = 2 (1)^{1/2} - 4 (1)^{1/4} + 4 ln ∣1 + 1∣ - 6

f (1) = 2 - 4 + 4 ln (2) - 6

The final result is:

$f (1) = 4 ln (2) - 8$

Common Traps & FAQs

Wait, why do we use

x^{1/4} = t

instead of just

x = t^{4}

directly?

That is a great question! While

x = t^{4}

is the same substitution, setting

t = x^{1/4}

is often more intuitive because it directly addresses the fractional power that makes the integral look complicated. By defining

t

as the root, we simplify the expression

x^{1/4}

t

immediately, making the substitution process cleaner and less prone to algebraic errors.

Why can't we just use partial fractions for

\frac{4 t ^{2}}{1 + t}

You absolutely can! Partial fractions are a powerful tool. However, for a simple rational function where the degree of the numerator is exactly one higher than the denominator, polynomial division or the 'add-and-subtract' trick (writing

t^{2}

t^{2} - 1 + 1

) is often faster and more elegant. It avoids setting up a system of equations and gets you straight to the integrable form.

Why do we reject the negative root or absolute value in the final log term?

In this specific problem, we have

t = x^{1/4}

. Since

x

is typically assumed to be non-negative in the context of real-valued functions with fractional powers like

x^{1/4}

t

will also be non-negative. Therefore,

1 + t

is always positive, which allows us to drop the absolute value bars in

ln ∣1 + t ∣

and write it simply as

ln (1 + t)

Animated Solution for Mathematics - Indefinite Integration: If $f (x) = \int \frac{1}{x ^{1/4} ( 1 + x ^{1/4} )} d x$ , $f (0) = - 6,$ then $f (1)$ is equal to:

Select Answer:

Visualized Solution

Substitution Strategy: $x^{\frac{1}{4}} = t$

Calculating the Differential $d x$

Transforming the Integral

Algebraic Manipulation of the Integrand

Integrating Term by Term

Back-substitution to $x$

Finding the Constant $C$

Calculating $f (1)$

Final Result and Conclusion

The Sigma Insight: Integration by Substitution

Analyzing the Setup

The Key of Substitution

The Algebraic Dance

The Calculus Core

Final Calculation

Similar Questions

If $f (x) = \int \frac{5 x ^{8} + 7 x ^{6}}{( x ^{2} + 1 + 2 x ^{7} ) ^{2}} d x, (x \geq 0)$ and $f (0) = 0$ , then the value of $f (1)$ is :

If $\int \frac{1}{x} \frac{1 - x}{1 + x} d x = g (x) + c, g (1) = 0$ , then $g (\frac{1}{2})$ is equal to :

Let $f (x) = \int x^{3} 3 - x^{2} d x$ . If $5 f (2) = - 4$ , then $f (1)$ is equal to

If $\int \frac{d x}{x ^{3} ( 1 + x ^{6} ) ^{2/3}} = x f (x) (1 + x^{6})^{1/3} + C$ where C is a constant of integration, then the function f(x) is equal to-

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Let $f (x) = \int \frac{d x}{x ^{(2/3)} + 2 x ^{(1/2)}}$ be such that $f (0) = - 26 + 24 lo g_{e} (2)$ . If $f (1) = a + b lo g_{e} (3)$ , where $a, b \in Z$ , then $a + b$ 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)

Evaluate $\int \frac{( x + 1 )}{x ( 1 + x e ^{x} ) ^{2}} d x$

Let $I (x) = \int \frac{6}{s i n ^{2} x ( 1 - c o t x ) ^{2}} d x$ . If $I (0) = 3$ , then $I (\frac{π}{12})$ is equal to

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Animated Solution for Mathematics - Indefinite Integration: .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } If f(x)=∫x1/4(1+x1/4)1​dx , f(0)=−6, then f(1) is equal to:

Select Answer:

Visualized Solution

Substitution Strategy: .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } x41​=t

Calculating the Differential .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } dx

Transforming the Integral

Algebraic Manipulation of the Integrand

Integrating Term by Term

Back-substitution to .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } x

Finding the Constant .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } C

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

Final Result and Conclusion

The Sigma Insight: Integration by Substitution

Analyzing the Setup

The Key of Substitution

The Algebraic Dance

The Calculus Core

Final Calculation

Similar Questions

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Animated Solution for Mathematics - Indefinite Integration: If $f (x) = \int \frac{1}{x ^{1/4} ( 1 + x ^{1/4} )} d x$ , $f (0) = - 6,$ then $f (1)$ is equal to:

Substitution Strategy: $x^{\frac{1}{4}} = t$

Calculating the Differential $d x$

Back-substitution to $x$

Finding the Constant $C$

Calculating $f (1)$

If $f (x) = \int \frac{5 x ^{8} + 7 x ^{6}}{( x ^{2} + 1 + 2 x ^{7} ) ^{2}} d x, (x \geq 0)$ and $f (0) = 0$ , then the value of $f (1)$ is :

If $\int \frac{1}{x} \frac{1 - x}{1 + x} d x = g (x) + c, g (1) = 0$ , then $g (\frac{1}{2})$ is equal to :

Let $f (x) = \int x^{3} 3 - x^{2} d x$ . If $5 f (2) = - 4$ , then $f (1)$ is equal to

If $\int \frac{d x}{x ^{3} ( 1 + x ^{6} ) ^{2/3}} = x f (x) (1 + x^{6})^{1/3} + C$ where C is a constant of integration, then the function f(x) is equal to-

Let $f (x) = \int \frac{x}{( 1 + x ) ^{2}} d x (x > 0)$ . Then $f (3) - f (1)$ is equal to :

Let $f (x) = \int \frac{d x}{x ^{(2/3)} + 2 x ^{(1/2)}}$ be such that $f (0) = - 26 + 24 lo g_{e} (2)$ . If $f (1) = a + b lo g_{e} (3)$ , where $a, b \in Z$ , then $a + b$ 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)

Evaluate $\int \frac{( x + 1 )}{x ( 1 + x e ^{x} ) ^{2}} d x$

Let $I (x) = \int \frac{6}{s i n ^{2} x ( 1 - c o t x ) ^{2}} d x$ . If $I (0) = 3$ , then $I (\frac{π}{12})$ is equal to

Let $f (x) = \int \frac{2 x}{( x ^{2} + 1 ) ( x ^{2} + 3 )} d x$ . If $f (3) = \frac{1}{2} (lo g_{e} 5 - lo g_{e} 6)$ , then $f (4)$ is equal to