JEE Main 2018 (15 April Shift 1)

LEVELJEE Main

Animated Solution for Mathematics - Indefinite Integration: 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)

Q: Wait, why can't I just integrate $2x + 1$ directly? Why do I have to find $f(x)$ first?

That is a classic trap! The expression $2x + 1$ is the output of the function $f$ when the input is $\frac{x-4}{x+2}$. It is not the value of $f(x)$. Integrating $2x+1$ would give you the integral of the output, not the integral of the function itself. You must map the input back to $x$ to find the true identity of $f(x)$ before you can perform the integration.

Select Answer:

Progressive Hints

Hint 1

The Spark: When you see a complex expression inside a function like

f (\frac{x - 4}{x + 2})

, do not panic. The secret is to treat the entire input as a single 'dummy variable',

t

. By setting

t = \frac{x - 4}{x + 2}

, you are essentially changing your coordinate system to see the function for what it truly is, rather than how it is disguised.

Hint 2

The Arsenal: Once you have your substitution

t = \frac{x - 4}{x + 2}

, you need to perform a 'variable inversion'. Your goal is to isolate

x

in terms of

t

. Use cross-multiplication, group all terms containing

x

on one side, and factor it out. This is a classic algebraic maneuver used to invert linear fractional transformations.

Hint 3

The Execution: After you derive

f (x) = \frac{3 x + 9}{1 - x}

, you face an improper rational function where the degree of the numerator equals the degree of the denominator. Do not attempt partial fractions immediately. Instead, manipulate the numerator to create the term

(1 - x)

, allowing you to split the integral into a simple constant and a logarithmic term.

Visualized Solution

Analyze the Functional Equation

Given functional equation: $f (\frac{x - 4}{x + 2}) = 2 x + 1$
Objective: Find $\int f (x) d x$
Constraint: $x \in R - {1, - 2}$

Substitution for Variable Change

Let $t = \frac{x - 4}{x + 2}$
We need to express $x$ as a function of $t$ .
This will allow us to find $f (t)$ .

Solving for $x$ in terms of $t$

$t = \frac{x - 4}{x + 2}$
Cross-multiply to eliminate the fraction:
$t (x + 2) = x - 4$
Expand the left side:
$t x + 2 t = x - 4$

Grouping $x$ Terms

$t x + 2 t = x - 4$
Move all terms containing $x$ to one side:
$t x - x = - 2 t - 4$
Factor out $x$ :
$x (t - 1) = - (2 t + 4)$

Final Expression for $x$

$x (t - 1) = - (2 t + 4)$
Divide by $(t - 1)$ :
$x = \frac{- ( 2 t + 4 )}{t - 1}$
Multiply numerator and denominator by $- 1$ :
$x = \frac{2 t + 4}{1 - t}$

Finding the Function $f (t)$

Recall the original right-hand side: $f (t) = 2 x + 1$
Substitute the expression for $x$ :
$f (t) = 2 (\frac{2 t + 4}{1 - t}) + 1$

Simplifying $f (t)$

$f (t) = \frac{2 ( 2 t + 4 )}{1 - t} + 1$
Take a common denominator:
$f (t) = \frac{4 t + 8 + ( 1 - t )}{1 - t}$
Combine like terms:
$f (t) = \frac{3 t + 9}{1 - t}$

Expressing $f (x)$

We have $f (t) = \frac{3 t + 9}{1 - t}$
Since $t$ is just a dummy variable, replace $t$ with $x$ :
$f (x) = \frac{3 x + 9}{1 - x}$

Preparing for Integration

Integral to solve: $I = \int \frac{3 x + 9}{1 - x} d x$
The degree of the numerator is equal to the degree of the denominator.
We need to manipulate the numerator to match the denominator $(1 - x)$ .

Rewriting the Numerator

Denominator is $1 - x$ .
Rewrite $3 x + 9$ in terms of $(1 - x)$ :
$3 x + 9 = - 3 (- x) + 9 = - 3 (1 - x - 1) + 9$
Alternatively: $3 x + 9 = - 3 (1 - x) + 3 + 9 = - 3 (1 - x) + 12$
Check: $- 3 + 3 x + 12 = 3 x + 9$ . Correct.

Splitting the Integral

Substitute the rewritten numerator into the integral:
$I = \int \frac{- 3 ( 1 - x ) + 12}{1 - x} d x$
Split the fraction into two parts:
$I = \int (\frac{- 3 ( 1 - x )}{1 - x} + \frac{12}{1 - x}) d x$
Simplify:
$I = \int (- 3 + \frac{12}{1 - x}) d x$

Execution of Integration

$I = \int - 3 d x + \int \frac{12}{1 - x} d x$
Integrate the first term: $\int - 3 d x = - 3 x$
Integrate the second term: $12 \int \frac{1}{1 - x} d x = 12 \frac{l o g _{e} ∣1 - x ∣}{- 1}$
Combine the results:
$I = - 3 x - 12 lo g_{e} ∣1 - x ∣ + C$

Final Answer and Conclusion

Final Result: $I = - 12 lo g_{e} ∣1 - x ∣ - 3 x + C$
Comparing with options, the correct option is Option 2.
Key Takeaway: Always find the explicit form of $f (x)$ first by using substitution in functional equations.

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The Sigma Insight: Integration by Substitution

Analyzing the Setup

The problem presents a functional equation:

f (\frac{x - 4}{x + 2}) = 2 x + 1

Many students mistakenly attempt to integrate the right-hand side directly. This is a trap; we must first isolate

f (x)

by performing a change of variables.

The Art of the Dummy Variable

We define a new variable

t

such that:

t = \frac{x - 4}{x + 2}

To express

f (t)

, we must solve for

x

in terms of

t

. Cross-multiplying gives:

t (x + 2) = x - 4

t x + 2 t = x - 4

Rearranging to isolate

x

t x - x = - 2 t - 4

x (t - 1) = - (2 t + 4)

x = \frac{2 t + 4}{1 - t}

Unmasking the Function

Now, we substitute this expression for

x

back into the original functional equation:

f (t) = 2 (\frac{2 t + 4}{1 - t}) + 1

To simplify, we find a common denominator:

f (t) = \frac{2 ( 2 t + 4 ) + ( 1 - t )}{1 - t}

f (t) = \frac{4 t + 8 + 1 - t}{1 - t} = \frac{3 t + 9}{1 - t}

Replacing the dummy variable

t

with

x

, we obtain the target function:

f (x) = \frac{3 x + 9}{1 - x}

The Calculus of Improper Fractions

We now evaluate the integral

I = \int f (x) d x

I = \int \frac{3 x + 9}{1 - x} d x

Since this is an improper rational function, we manipulate the numerator to match the denominator:

3 x + 9 = - 3 (1 - x) + 12

Substituting this back into the integral:

I = \int \frac{- 3 ( 1 - x ) + 12}{1 - x} d x

I = \int - 3 d x + \int \frac{12}{1 - x} d x

The Final Victory

Integrating the two terms separately, we account for the chain rule in the second term:

I = - 3 x - 12 ln ∣1 - x ∣ + C

The final result is $- 3 x - 12 ln ∣1 - x ∣ + C$ . This systematic approach ensures that even complex functional equations are reduced to standard, solvable forms.

Common Traps & FAQs

Wait, why can't I just integrate

2 x + 1

directly? Why do I have to find

f (x)

first?

That is a classic trap! The expression

2 x + 1

is the output of the function

f

when the input is

\frac{x - 4}{x + 2}

. It is not the value of

f (x)

. Integrating

2 x + 1

would give you the integral of the output, not the integral of the function itself. You must map the input back to

x

to find the true identity of

f (x)

before you can perform the integration.

Why did the integral of

\frac{12}{1 - x}

result in

- 12 lo g_{e} ∣1 - x ∣

? Where did the negative sign come from?

Excellent observation! This is where the Chain Rule comes into play. When you integrate

\frac{1}{u}

, you get

ln ∣ u ∣

. However, because your denominator is

(1 - x)

, you must divide by the derivative of the inner function, which is

\frac{d}{d x} (1 - x) = - 1

. That division by

- 1

is exactly what flips the sign. Always watch your signs when the coefficient of

x

is negative!

Is there a faster way to rewrite

\frac{3 x + 9}{1 - x}

without doing all that algebra?

You can use the 'Long Division' method or the 'Adjustment' method. The adjustment method I showed you—forcing the numerator to look like the denominator—is generally faster and less prone to calculation errors than long division. It builds your algebraic intuition, which is a superpower in JEE Advanced. Keep practicing this 'make-it-look-like-the-denominator' trick; it will save you minutes in the exam hall.

Animated Solution for Mathematics - Indefinite Integration: 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)

Select Answer:

Visualized Solution

Analyze the Functional Equation

Substitution for Variable Change

Solving for $x$ in terms of $t$

Grouping $x$ Terms

Final Expression for $x$

Finding the Function $f (t)$

Simplifying $f (t)$

Expressing $f (x)$

Preparing for Integration

Rewriting the Numerator

Splitting the Integral

Execution of Integration

Final Answer and Conclusion

The Sigma Insight: Integration by Substitution

Analyzing the Setup

The Art of the Dummy Variable

Unmasking the Function

The Calculus of Improper Fractions

The Final Victory

Similar Questions

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:

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

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

If $\int (e^{2 x} + 2 e^{x} - e^{- x} - 1) e^{(e^{x} + e^{- x})} d x = g (x) e^{(e^{x} + e^{- x})} + c$ , where $c$ is a constant of integration, then $g (0)$ 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-

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

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

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 :

If $\int \frac{c o s x d x}{s i n ^{2} x ( 1 + s i n x ) ^{2/3}} = f (x) (1 + sin^{6} x)^{1/ λ} + c$ where $c$ is a constant of integration, then $λ f (\frac{π}{3})$ is equal to :

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

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+2x−4​)=2x+1,(x∈R−{1,−2}), then ∫f(x)dx is equal to : (where C is a constant of integration)

Select Answer:

Visualized Solution

Analyze the Functional Equation

Substitution for Variable Change

Solving for .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } x in terms of .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } t

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

Final Expression for .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } x

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

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

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

Preparing for Integration

Rewriting the Numerator

Splitting the Integral

Execution of Integration

Final Answer and Conclusion

The Sigma Insight: Integration by Substitution

Analyzing the Setup

The Art of the Dummy Variable

Unmasking the Function

The Calculus of Improper Fractions

The Final Victory

Similar Questions

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.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Evaluate ∫x(1+xex)2(x+1)​dx

Animated Solution for Mathematics - Indefinite Integration: 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)

Solving for $x$ in terms of $t$

Grouping $x$ Terms

Final Expression for $x$

Finding the Function $f (t)$

Simplifying $f (t)$

Expressing $f (x)$

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:

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

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

If $\int (e^{2 x} + 2 e^{x} - e^{- x} - 1) e^{(e^{x} + e^{- x})} d x = g (x) e^{(e^{x} + e^{- x})} + c$ , where $c$ is a constant of integration, then $g (0)$ 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-

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

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

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 :

If $\int \frac{c o s x d x}{s i n ^{2} x ( 1 + s i n x ) ^{2/3}} = f (x) (1 + sin^{6} x)^{1/ λ} + c$ where $c$ is a constant of integration, then $λ f (\frac{π}{3})$ is equal to :

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