JEE Main 2025 (January)

LEVELJEE Main

Animated Solution for Mathematics - Indefinite Integration: If $\int \frac{2 x ^{2} + 5 x + 9}{x ^{2} + x + 1} d x = x x^{2} + x + 1 + α x^{2} + x + 1 + β lo g_{e} ∣ x + \frac{1}{2} + x^{2} + x + 1 ∣ + C .$ where C is the constant of integration, then $α + 2 β$ is equal to

Q: Wait, why can't we just use substitution $u = x^{2}+x+1$ directly?

That is a great question! If the numerator were simply $2x+1$, then $u$-substitution would be perfect. However, because the numerator is a full quadratic ($2x^{2}+5x+9$), a simple substitution won't eliminate the $x^{2}$ term. We must decompose the numerator first to isolate the part that matches the derivative of the denominator.

Q: How do I know the decomposition is $A(Q) + B(Q') + C$?

This is a standard technique for integrals of the form $\int \frac{Px^{2}+Qx+R}{\sqrt{ax^{2}+bx+c}} dx$. By using $A(Q) + B(Q') + C$, we create three distinct types of integrals: one that simplifies to $\sqrt{Q}$, one that is $\frac{Q'}{\sqrt{Q}}$ (which is easily integrable via substitution), and one that is $\frac{1}{\sqrt{Q}}$ (which is a standard integral form).

Q: Why do we get a log term in the final answer?

The log term arises from the third integral, $\int \frac{1}{\sqrt{x^{2}+x+1}} dx$. After completing the square in the denominator, it takes the form $\int \frac{1}{\sqrt{(x+1/2)^{2} + 3/4}} dx$. This is a standard integral that evaluates to $\ln|x+1/2 + \sqrt{x^{2}+x+1}|$. It is a fundamental result in calculus that you should keep in your toolkit!

Enter Numerical Value:

Progressive Hints

Hint 1

Hint 1 (The Spark): Visualize the structure of the integral. You are dealing with a quadratic polynomial divided by the square root of another quadratic. The key is to decompose the numerator into parts that directly relate to the denominator and its derivative, effectively 'splitting' the monster into three manageable pieces.

Hint 2

Hint 2 (The Arsenal): Use the method of numerator decomposition. Express the numerator

2 x^{2} + 5 x + 9

A (x^{2} + x + 1) + B \frac{d}{d x} (x^{2} + x + 1) + C

. This allows you to transform the integral into a sum of three simpler integrals: one that simplifies to a square root, one that is a standard substitution, and one that is a standard logarithmic form.

Hint 3

Hint 3 (The Execution): Start by setting up the identity:

2 x^{2} + 5 x + 9 = A (x^{2} + x + 1) + B (2 x + 1) + C

. Solve for

A

B

, and

C

by comparing coefficients of

x^{2}

x

, and the constant term. Once you have these, substitute them back into the integral and evaluate each part separately.

Visualized Solution

Analyze the Integral Form

Given Integral: $I = \int \frac{2 x ^{2} + 5 x + 9}{x ^{2} + x + 1} d x$
Target Form: $x x^{2} + x + 1 + α x^{2} + x + 1 + β lo g_{e} ∣ x + \frac{1}{2} + x^{2} + x + 1 ∣ + C$
Objective: Find the value of $α + 2 β$ .

Numerator Decomposition Strategy

Let $2 x^{2} + 5 x + 9 = A (x^{2} + x + 1) + B \frac{d}{d x} (x^{2} + x + 1) + C$
Derivative of denominator term: $\frac{d}{d x} (x^{2} + x + 1) = 2 x + 1$
Identity: $2 x^{2} + 5 x + 9 = A (x^{2} + x + 1) + B (2 x + 1) + C$

Finding Coefficient $A$

Comparing coefficients of $x^{2}$ on both sides:
Left Hand Side: $2$
Right Hand Side: $A$
Therefore, $A = 2$

Finding Coefficient $B$

Comparing coefficients of $x$ on both sides:
Left Hand Side: $5$
Right Hand Side: $A + 2 B$
Substitute $A = 2$ : $2 + 2 B = 5 \Rightarrow 2 B = 3 \Rightarrow B = \frac{3}{2}$

Finding Constant $C$

Comparing constant terms on both sides:
Left Hand Side: $9$
Right Hand Side: $A + B + C$
Substitute $A = 2, B = \frac{3}{2}$ :
$2 + \frac{3}{2} + C = 9 \Rightarrow \frac{7}{2} + C = 9$
$C = 9 - \frac{7}{2} = \frac{11}{2}$

Rewriting the Integral

Substitute $A, B, C$ back into the integral:
$I = \int \frac{2 ( x ^{2} + x + 1 ) + \frac{3}{2} ( 2 x + 1 ) + \frac{11}{2}}{x ^{2} + x + 1} d x$
Split into three integrals:
$I = 2 \int x^{2} + x + 1 d x + \frac{3}{2} \int \frac{2 x + 1}{x ^{2} + x + 1} d x + \frac{11}{2} \int \frac{1}{x ^{2} + x + 1} d x$

Solving Part $1$

Complete the square: $x^{2} + x + 1 = (x + \frac{1}{2})^{2} + \frac{3}{4}$
Use $\int x^{2} + a^{2} d x = \frac{x}{2} x^{2} + a^{2} + \frac{a ^{2}}{2} ln ∣ x + x^{2} + a^{2} ∣$
$2 \int (x + \frac{1}{2})^{2} + \frac{3}{4} d x = 2 [\frac{x + \frac{1}{2}}{2} x^{2} + x + 1 + \frac{\frac{3}{4}}{2} ln ∣ x + \frac{1}{2} + x^{2} + x + 1 ∣]$
$= (x + \frac{1}{2}) x^{2} + x + 1 + \frac{3}{4} ln ∣ x + \frac{1}{2} + x^{2} + x + 1 ∣$

Solving Part $2$

Integral: $\frac{3}{2} \int \frac{2 x + 1}{x ^{2} + x + 1} d x$
Let $u = x^{2} + x + 1 \Rightarrow d u = (2 x + 1) d x$
$= \frac{3}{2} \int u^{- \frac{1}{2}} d u = \frac{3}{2} (2 u^{\frac{1}{2}}) = 3 x^{2} + x + 1$

Solving Part $3$

Integral: $\frac{11}{2} \int \frac{1}{x ^{2} + x + 1} d x$
Use $\int \frac{1}{x ^{2} + a ^{2}} d x = ln ∣ x + x^{2} + a^{2} ∣$
$= \frac{11}{2} ln ∣ x + \frac{1}{2} + x^{2} + x + 1 ∣$

Aggregating All Terms

Combine all results:
$I = (x + \frac{1}{2}) x^{2} + x + 1 + 3 x^{2} + x + 1 + (\frac{3}{4} + \frac{11}{2}) ln ∣ x + \frac{1}{2} + x^{2} + x + 1 ∣ + C$
Simplify root terms: $(x + \frac{1}{2} + 3) x^{2} + x + 1 = (x + \frac{7}{2}) x^{2} + x + 1$
Simplify log terms: $\frac{3}{4} + \frac{22}{4} = \frac{25}{4}$

Comparing to Find $α$ and $β$

Result: $x x^{2} + x + 1 + \frac{7}{2} x^{2} + x + 1 + \frac{25}{4} ln ∣ x + \frac{1}{2} + x^{2} + x + 1 ∣ + C$
Given form: $x x^{2} + x + 1 + α x^{2} + x + 1 + β lo g_{e} ∣ x + \frac{1}{2} + x^{2} + x + 1 ∣ + C$
By comparison: $α = \frac{7}{2}$ and $β = \frac{25}{4}$

Final Calculation

Calculate $α + 2 β$ :
$α + 2 β = \frac{7}{2} + 2 (\frac{25}{4})$
$= \frac{7}{2} + \frac{25}{2} = \frac{32}{2}$
Final Answer: $16$

Summary and Takeaways

Key Takeaway: Use the form $P x^{2} + Q x + R = A (a x^{2} + b x + c) + B (2 a x + b) + C$ for integrals involving $\frac{Quadratic}{Quadratic}$ .
Next Challenge: Try solving the same integral if the numerator was a cubic expression! How would the decomposition change?

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The Sigma Insight: Evaluation of Special Integral Forms

The Anatomy of a Monster Integral

Welcome, fellow traveler on the path to JEE mastery. Today, we face a problem that might look like a chaotic mess of algebra at first glance:

\int \frac{2 x ^{2} + 5 x + 9}{x ^{2} + x + 1} d x

When you see a quadratic expression sitting atop the square root of another quadratic, your instinct might be to panic. But take a deep breath. In the world of JEE Advanced, every 'monster' has a weakness. Our weakness today is a surgical technique called Numerator Decomposition.

Phase 1

The Surgical Decomposition

We cannot integrate this directly. The numerator is too heavy. We need to break it down into pieces that the denominator can 'understand.'

We want to write the numerator

2 x^{2} + 5 x + 9

as a combination of three distinct parts: 1. The quadratic itself:

(x^{2} + x + 1)

2. The derivative of the quadratic:

\frac{d}{d x} (x^{2} + x + 1) = 2 x + 1

3. A constant:

C

So, we set up the identity:

2 x^{2} + 5 x + 9 = A (x^{2} + x + 1) + B (2 x + 1) + C

Now, we play the matching game. By comparing the coefficients of

x^{2}

x

, and the constant terms on both sides, we find our constants.

Comparing

x^{2}

gives

A = 2

. Comparing

x

gives

A + 2 B = 5

, which, with

A = 2

, leads to

2 B = 3

, or

B = \frac{3}{2}

Finally, comparing the constants gives

A + B + C = 9

, which means

2 + \frac{3}{2} + C = 9

, resulting in

C = \frac{11}{2}

Phase 2

The Three-Fold Path

With

A = 2

B = \frac{3}{2}

, and

C = \frac{11}{2}

, our integral splits into three beautiful, manageable parts:

I = 2 \int x^{2} + x + 1 d x + \frac{3}{2} \int \frac{2 x + 1}{x ^{2} + x + 1} d x + \frac{11}{2} \int \frac{1}{x ^{2} + x + 1} d x

Look at what we have achieved! The first part is a standard square root integral. The second part is a simple substitution (

u = x^{2} + x + 1

), and the third is a standard logarithmic form. We have tamed the beast.

Phase 3

The Final Assembly

Solving these individually, we apply the standard formula for

\int x^{2} + a^{2} d x

for the first part, the power rule for the second, and the logarithmic integral for the third. After careful calculation and combining the terms, we arrive at our result:

I = (x + \frac{1}{2}) x^{2} + x + 1 + 3 x^{2} + x + 1 + \frac{25}{4} ln ∣ x + \frac{1}{2} + x^{2} + x + 1 ∣ + C

Simplifying the root terms, we get

(x + \frac{7}{2}) x^{2} + x + 1

. Comparing this to the target form

x x^{2} + x + 1 + α x^{2} + x + 1 + β ln ∣ x + \frac{1}{2} + x^{2} + x + 1 ∣ + C

, we identify

α = \frac{7}{2}

and

β = \frac{25}{4}

The Victory Lap

Finally, we compute

α + 2 β

α + 2 β = \frac{7}{2} + 2 (\frac{25}{4}) = \frac{7}{2} + \frac{25}{2} = \frac{32}{2} = 16

There it is! The monster is defeated. Remember, the complexity of an integral is often just a test of your patience and your ability to decompose the problem into smaller, solvable truths. Keep practicing, and keep that mathematical fire burning! The final answer is 16.

Common Traps & FAQs

Wait, why can't we just use substitution

u = x^{2} + x + 1

directly?

That is a great question! If the numerator were simply

2 x + 1

, then

u

-substitution would be perfect. However, because the numerator is a full quadratic (

2 x^{2} + 5 x + 9

), a simple substitution won't eliminate the

x^{2}

term. We must decompose the numerator first to isolate the part that matches the derivative of the denominator.

How do I know the decomposition is

A (Q) + B (Q^{'}) + C

This is a standard technique for integrals of the form

\int \frac{P x ^{2} + Q x + R}{a x ^{2} + b x + c} d x

. By using

A (Q) + B (Q^{'}) + C

, we create three distinct types of integrals: one that simplifies to

Q

, one that is

\frac{Q ^{'}}{Q}

(which is easily integrable via substitution), and one that is

\frac{1}{Q}

(which is a standard integral form).

Why do we get a log term in the final answer?

The log term arises from the third integral,

\int \frac{1}{x ^{2} + x + 1} d x

. After completing the square in the denominator, it takes the form

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

. This is a standard integral that evaluates to

ln ∣ x + 1/2 + x^{2} + x + 1 ∣

. It is a fundamental result in calculus that you should keep in your toolkit!

Animated Solution for Mathematics - Indefinite Integration: If $\int \frac{2 x ^{2} + 5 x + 9}{x ^{2} + x + 1} d x = x x^{2} + x + 1 + α x^{2} + x + 1 + β lo g_{e} ∣ x + \frac{1}{2} + x^{2} + x + 1 ∣ + C .$ where C is the constant of integration, then $α + 2 β$ is equal to

Enter Numerical Value:

Visualized Solution

Analyze the Integral Form

Numerator Decomposition Strategy

Finding Coefficient $A$

Finding Coefficient $B$

Finding Constant $C$

Rewriting the Integral

Solving Part $1$

Solving Part $2$

Solving Part $3$

Aggregating All Terms

Comparing to Find $α$ and $β$

Final Calculation

Summary and Takeaways

The Sigma Insight: Evaluation of Special Integral Forms

The Anatomy of a Monster Integral

Phase 1

Phase 2

Phase 3

The Victory Lap

Similar Questions

If $\int \frac{d x}{( x ^{2} + x + 1 ) ^{2}} = a tan^{- 1} (\frac{2 x + 1}{3}) + b (\frac{2 x + 1}{x ^{2} + x + 1}) + C, x > 0$ where $C$ is the constant of integration, then the value of $9 (3 a + b)$ is equal to.

If $\int \frac{c o s θ}{5 + 7 s i n θ - 2 c o s ^{2} θ} d θ = A lo g_{e} ∣ B (θ) ∣ + C$ where $C$ is a constant of integration, then $\frac{B ( θ )}{A}$ can be:

The integral $\int ((\frac{x}{2})^{x} + (\frac{2}{x})^{x}) lo g_{2} x d x$ is equal to

If $\int e^{x} (\frac{x s i n ^{- 1} x}{1 - x ^{2}} + \frac{s i n ^{- 1} x}{( 1 - x ^{2} ) ^{3/2}} + \frac{x}{1 - x ^{2}}) d x = g (x) + C,$ where C is the constant of integration, then $g (\frac{1}{2})$ equals:

If $\int \frac{5 t a n x}{t a n x - 2} d x = x + a ln ∣ sin x - 2 cos x ∣ + k$ , then $a$ is equal to

Let $f (t) = \int (\frac{1 - s i n ( l o g _{e} t )}{1 - c o s ( l o g _{e} t )}) d t, t > 1$ . If $f (e^{π /2}) = - e^{π /2}$ and $f (e^{π /4}) = α e^{π /4}$ , then $α$ equals

If $\int \frac{d x}{( x ^{2} - 2 x + 10 ) ^{2}} = A (tan^{- 1} (\frac{x - 1}{3}) + \frac{f ( x )}{x ^{2} - 2 x + 10}) + C$ , where $C$ is a constant of integration, then :

Let $I = \int \frac{e ^{x}}{e ^{4 x} + e ^{2 x} + 1} d x, J = \int \frac{e ^{- x}}{e ^{- 4 x} + e ^{- 2 x} + 1} d x$ . Then, for an arbitrary constant $C$ , the value of $J - I$ equals

Enter Numerical Value:

Visualized Solution

Analyze the Integral Form

Numerator Decomposition Strategy

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

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

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

Rewriting the Integral

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

Solving Part .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } 2

Solving Part .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } 3

Aggregating All Terms

Comparing to Find .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } α and .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } β

Final Calculation

Summary and Takeaways

The Sigma Insight: Evaluation of Special Integral Forms

The Anatomy of a Monster Integral

Phase 1

Phase 2

Phase 3

The Victory Lap

Similar Questions

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } If ∫(x2+x+1)2dx​=atan−1(3​2x+1​)+b(x2+x+12x+1​)+C,x>0 where C is the constant of integration, then the value of 9(3​a+b) is equal to.

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } If ∫5+7sinθ−2cos2θcosθ​dθ=Aloge​∣B(θ)∣+C where C is a constant of integration, then AB(θ)​ can be:

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } The integral ∫((2x​)x+(x2​)x)log2​xdx is equal to

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } If ∫ex(1−x2​x sin−1x​+(1−x2)3/2sin−1x​+1−x2x​)dx=g(x)+C, where C is the constant of integration, then g(21​) equals:

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } If ∫tanx−25tanx​dx=x+aln∣sinx−2cosx∣+k, then a is equal to

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Let f(t)=∫(1−cos(loge​t)1−sin(loge​t)​)dt,t>1. If f(eπ/2)=−eπ/2 and f(eπ/4)=αeπ/4, then α equals

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } If ∫(x2−2x+10)2dx​=A(tan−1(3x−1​)+x2−2x+10f(x)​)+C, where C is a constant of integration, then :

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Let I=∫e4x+e2x+1ex​dx,J=∫e−4x+e−2x+1e−x​dx. Then, for an arbitrary constant C, the value of J−I equals

Finding Coefficient $A$

Finding Coefficient $B$

Finding Constant $C$

Solving Part $1$

Solving Part $2$

Solving Part $3$

Comparing to Find $α$ and $β$

If $\int \frac{d x}{( x ^{2} + x + 1 ) ^{2}} = a tan^{- 1} (\frac{2 x + 1}{3}) + b (\frac{2 x + 1}{x ^{2} + x + 1}) + C, x > 0$ where $C$ is the constant of integration, then the value of $9 (3 a + b)$ is equal to.

If $\int \frac{c o s θ}{5 + 7 s i n θ - 2 c o s ^{2} θ} d θ = A lo g_{e} ∣ B (θ) ∣ + C$ where $C$ is a constant of integration, then $\frac{B ( θ )}{A}$ can be:

The integral $\int ((\frac{x}{2})^{x} + (\frac{2}{x})^{x}) lo g_{2} x d x$ is equal to

If $\int e^{x} (\frac{x s i n ^{- 1} x}{1 - x ^{2}} + \frac{s i n ^{- 1} x}{( 1 - x ^{2} ) ^{3/2}} + \frac{x}{1 - x ^{2}}) d x = g (x) + C,$ where C is the constant of integration, then $g (\frac{1}{2})$ equals:

If $\int \frac{5 t a n x}{t a n x - 2} d x = x + a ln ∣ sin x - 2 cos x ∣ + k$ , then $a$ is equal to

Let $f (t) = \int (\frac{1 - s i n ( l o g _{e} t )}{1 - c o s ( l o g _{e} t )}) d t, t > 1$ . If $f (e^{π /2}) = - e^{π /2}$ and $f (e^{π /4}) = α e^{π /4}$ , then $α$ equals

If $\int \frac{d x}{( x ^{2} - 2 x + 10 ) ^{2}} = A (tan^{- 1} (\frac{x - 1}{3}) + \frac{f ( x )}{x ^{2} - 2 x + 10}) + C$ , where $C$ is a constant of integration, then :

Let $I = \int \frac{e ^{x}}{e ^{4 x} + e ^{2 x} + 1} d x, J = \int \frac{e ^{- x}}{e ^{- 4 x} + e ^{- 2 x} + 1} d x$ . Then, for an arbitrary constant $C$ , the value of $J - I$ equals