60pi|(3x+2x+ x)|dx is equal to..... — jee_main_2026_21_jan

Q72 jee_main_2025_07_april_evening Integration by Substitution

If

\int ((1)/(x) +(1)/(x³))( [2]3x⁻²⁴ + x⁻²⁶)dx = - (α)/(3 (α + 1)) (3 x ^ β + x ^ γ) ^ (α + 1)/(α) + C

,

x > 0, (α, β, γ in Z)

, where

C

is the constant of integration, then

α + β + γ

is equal to

Numerical Answer. Answer: 19 to 19

Solution

Related Formula

Standard power integration formula rule is:

\int uⁿ du = uⁿ⁺¹n+1 + C

Core Logic

Rewrite the integral by adjusting powers inside the radical container:

I = \int ((1)/(x²) + (1)/(x⁴)) ((3)/(x) + (1)/(x³))(1)/(23) dx

Let $t = (3)/(x) + (1)/(x³) dt = -3((1)/(x²) + (1)/(x⁴)) dx$ .

Step 1: Integrate

Substituting $t$ $t$ into the equation:

\int t1/23 dt-3 = -(1)/(3) \cdot t24/2324/23 = -(23)/(3(24)) t(24)/(23)

Comparing parameters directly yields:

α = 23, β = -1, γ = -3

α + β + γ = 23 - 1 - 3 = 19

Pattern Recognition

Pull out high powers from the root factor to seamlessly reveal a matching $f'(x)$ $f'(x)$ substitution pair on the outside.

Chapter Mix

Class 12 Mathematics: Integral Calculus

Q75 jee_main_2025_24_jan_evening Indefinite Integration of Algebraic Functions

If

\int 2x²+5x+9 x²+x+1dx=x x²+x+1+α x²+x+1+β ₑ|x+(1)/(2)+ x²+x+1|+C

where C is the constant of integration, then

α+2β

is equal to \_\_\_\_.

Numerical Answer. Answer: 16

Solution

Related Formula

Standard integration templates for quadratic forms:

\int \sqrt(t² + k²) dt = (t)/(2)\sqrt(t²+k²) + (k²)/(2)ln|t + \sqrt(t²+k²)|

\int 1\sqrt(t² + k²) dt = ln|t + \sqrt(t²+k²)|

Core Logic

Decompose the numerator using polynomial differentiation components :

2x² + 5x + 9 = A(x² + x + 1) + B(2x + 1) + C

Equating coefficients dynamically yields :

For $x²$ $x^2$: $A = 2$ $A = 2$ .
For $x$ $x$: $A + 2B = 5 \Rightarrow 2 + 2B = 5 \Rightarrow B = (3)/(2)$ .
Constant: $A + B + C = 9 \Rightarrow 2 + (3)/(2) + C = 9 \Rightarrow C = (11)/(2)$ .

Rewrite the integrand into three parts :

2\int \sqrt(x²+x+1) dx + (3)/(2)\int 2x+1\sqrt(x²+x+1) dx + (11)/(2)\int 1\sqrt(x²+x+1) dx

Step 1: Complete Quadratics & Integrate

Format using completing the square technique: $x² + x + 1 = (x + (1)/(2))² + ( \sqrt(3)2)²$ .

First Integral evaluation:

2 [ ((x+(1)/(2)))/(2)\sqrt(x²+x+1) + (3)/(8)ln|x+(1)/(2)+\sqrt(x²+x+1)| ]

= (x+(1)/(2))\sqrt(x²+x+1) + (3)/(4)ln|x+(1)/(2)+\sqrt(x²+x+1)|

Second Integral evaluation :

(3)/(2) \cdot 2\sqrt(x²+x+1) = 3\sqrt(x²+x+1)

Third Integral evaluation :

(11)/(2)ln|x+(1)/(2)+\sqrt(x²+x+1)|

Step 2: Collect Like Terms

Gather and combine matching factor parameters:

Total = (x + (1)/(2) + 3)\sqrt(x²+x+1) + ((3)/(4) + (11)/(2))ln|x+(1)/(2)+\sqrt(x²+x+1)|

Total = (x + (7)/(2))\sqrt(x²+x+1) + (25)/(4)ln|x+(1)/(2)+\sqrt(x²+x+1)|

= x\sqrt(x²+x+1) + (7)/(2)\sqrt(x²+x+1) + (25)/(4)ln|x+(1)/(2)+\sqrt(x²+x+1)|

Step 3: Extract Coefficients

Compare directly against the given expression variables :

α = (7)/(2), β = (25)/(4)

α + 2β = (7)/(2) + 2((25)/(4)) = (7)/(2) + (25)/(2) = (32)/(2) = 16

Pattern Recognition

When dividing large numerators containing $x²$ $x^2$ elements over quadratic square roots, using matching coefficient expansion rules prevents lengthy substitution errors completely.

Chapter Mix

Class 12 Mathematics: Integrals

Q52 jee_main_2025_24_jan_morning Properties of Definite Integrals

If

I(m,n) = \int₀¹ xm-1 (1-x)ⁿ⁻¹ dx

where

m, n > 0

, then

I(9,14) + I(10,13)

is :

A. $I(9, 1)$ $I(9, 1)$
B. $I(19, 27)$ $I(19, 27)$
C. $I(1, 13)$ $I(1, 13)$
D. $I(9, 13)$ $I(9, 13)$

Solution

Related Formula

The beta function integral format satisfies:

I(m,n) = \int₀¹ xm-1 (1-x)ⁿ⁻¹ dx

Core Logic

Let's combine the terms of the requested sum directly by inserting their respective definitions:

I(9,14) = \int₀¹ x⁹⁻¹ (1-x)¹⁴⁻¹ dx = \int₀¹ x⁸ (1-x)¹³ dx

I(10,13) = \int₀¹ x¹⁰⁻¹ (1-x)¹³⁻¹ dx = \int₀¹ x⁹ (1-x)¹² dx

Step 1: Factoring out common algebraic terms

Summing the two components:

I(9,14) + I(10,13) = \int₀¹ [ x⁸ (1-x)¹³ + x⁹ (1-x)¹² ] dx

Factor out the common term $x⁸ (1-x)¹²$ $x^8 (1-x)^{12}$ inside the integrand:

= \int₀¹ x⁸ (1-x)¹² [ (1-x) + x ] dx

= \int₀¹ x⁸ (1-x)¹² (1) dx

= \int₀¹ x⁹⁻¹ (1-x)¹³⁻¹ dx = I(9,13)

Pattern Recognition

When dealing with linear combinations of beta functions with shifting parameter indices, directly writing down the definite integral expression often results in immediate algebraic cancellation or simplification via basic factoring.

Chapter Mix

Class 12 Mathematics: Definite Integrals

Q71 jee_main_2025_24_jan_morning Differentiating Under the Integral Sign

Let

f

be a differentiable function such that

2(x+2)²f(x) - 3(x+2)² = 10\int₀x(t+2)f(t)dt

for

x \geq 0

. Then

f(2)

is equal to ________.

Numerical Answer. Answer: 19

Solution

Related Formula

The Leibniz Integral Rule template allows direct differentiation of an integral with variable limits:

(d)/(dx)( \int₀x g(t) dt ) = g(x)

Core Logic

Differentiate both sides of the given functional equation with respect to $x$ $x$ using the product rule:

(d)/(dx)[ 2(x+2)² f(x) - 3(x+2)² ] = (d)/(dx)[ 10\int₀x(t+2)f(t)dt ]

4(x+2)f(x) + 2(x+2)² f'(x) - 6(x+2) = 10(x+2)f(x)

Since $x \geq 0$ $x \geq 0$, the factor $(x+2)$ $(x+2)$ is strictly non-zero. Divide the entire equation by $2(x+2)$ $2(x+2)$:

2f(x) + (x+2)f'(x) - 3 = 5f(x)

(x+2)f'(x) - 3f(x) = 3

Step 1: Solve the First-Order Differential Equation

Rearrange the expression into standard linear differential equation form where $y = f(x)$ $y = f(x)$:

(dy)/(dx) - (3)/(x+2)y = (3)/(x+2)

Compute the Integrating Factor (I.F.):

I.F. = e\int -(3)/(x+2) dx = e-3ln(x+2) = (x+2)⁻³

Multiply through by the I.F. and integrate:

y \cdot (x+2)⁻³ = \int (3)/(x+2) \cdot (x+2)⁻³ dx = \int 3(x+2)⁻⁴ dx

(f(x))/((x+2)³) = 3 \cdot (x+2)⁻³-3 + C = -(x+2)⁻³ + C

f(x) = -1 + C(x+2)³

Step 2: Apply the Boundary Condition

Find the boundary condition by substituting $x = 0$ $x = 0$ into the original integral equation equation:

2(0+2)² f(0) - 3(0+2)² = 10 \int₀⁰ (t+2)f(t) dt

8f(0) - 12 = 0 f(0) = (12)/(8) = (3)/(2)

Substitute $x = 0$ $x = 0$ into our general solution formula:

f(0) = -1 + C(0+2)³ (3)/(2) = -1 + 8C

(5)/(2) = 8C C = (5)/(16)

Thus, the explicit function is:

f(x) = -1 + (5)/(16)(x+2)³

Step 3: Evaluate at target point x = 2

Substitute $x = 2$ $x = 2$ into the final function equation:

f(2) = -1 + (5)/(16)(2+2)³ = -1 + (5)/(16)(64)

f(2) = -1 + 5(4) = -1 + 20 = 19

Pattern Recognition

When an equation contains a variable integral limit $\int₀^x$ $\int_0^x$, differentiating both sides using the Leibniz rule converts it into a standard differential equation. The initial value is found by setting $x = 0$ $x = 0$ directly in the original expression.

Chapter Mix

Class 12 Mathematics: Definite Integrals Class 12 Mathematics: Differential Equations

Q jee_main_2025_28_jan_evening Integration by Substitution

If

f(x) = \int 1x1/4(1+x1/4) dx

,

f(0) = -6

, then

f(1)

is equal to:

A. $ₑ2+2$ $\log_{e}2+2$
B. $4( ₑ2-2)$ $4(\log_{e}2-2)$
C. $2- ₑ2$ $2-\log_{e}2$
D. $4( ₑ2+2)$ $4(\log_{e}2+2)$

Solution

Related Formula

Standard substitution method and logarithmic integral rule:

\int (1)/(t+1) dt = ln|t+1| + C

Core Logic

Let $x = t⁴ dx = 4t³ dt$ $x = t^4 \implies dx = 4t^3 dt$. When substituting into the integral:

f(x) = \int (4t³)/(t(1+t)) dt = 4 \int (t²)/(1+t) dt

Step 1: Simplify the Integral

Rewrite the numerator $t²$ $t^2$ as $(t² - 1) + 1$ $(t^2 - 1) + 1$:

4 \int ((t² - 1) + 1)/(1+t) dt = 4 \int ( ((t-1)(t+1))/(1+t) + (1)/(1+t) ) dt

4 \int (t - 1) dt + 4 \int (1)/(t+1) dt

4 [ ((t-1)²)/(2) ] + 4 ln|t+1| + C = 2(t-1)² + 4 ln|t+1| + C

Substitute back $t = x1/4$ $t = x^{1/4}$:

f(x) = 2(x1/4 - 1)² + 4 ln(1 + x1/4) + C

Step 2: Solve for Constant C and Find f(1)

Given $f(0) = -6$ $f(0) = -6$:

-6 = 2(0 - 1)² + 4 ln(1 + 0) + C

-6 = 2 + 0 + C C = -8

Now find $f(1)$ $f(1)$:

f(1) = 2(11/4 - 1)² + 4 ln(1 + 11/4) - 8

f(1) = 2(0) + 4 ln(2) - 8 = 4 ln 2 - 8 = 4(ln 2 - 2)

Pattern Recognition

By adding and subtracting terms in the numerator ( $t²-1+1$ $t^2-1+1$), we can quickly bypass long division for polynomials and directly integrate using standard forms.

Chapter Mix

Class 12 Mathematics: Indefinite Integration

6∫₀π|( 3x+ 2x+ x)|dx$6\int_{0}^{\pi}\left|\left(\sin3x+\sin2x+\sin x\right)\right|dx$ is equal to.....

Solution & Explanation

Related Formula

Core Logic

Step 1: Coordinate Substitution

Step 2: Resolve Modulus Intervals

Step 3: Evaluate Integrals

Step 4: Final Output

Pattern Recognition

Chapter Mix

Reference Study Guides

More Definite Integration Previous-Year Questions — Page 8

Solution

Related Formula

Core Logic

Step 1: Integrate

Pattern Recognition

Chapter Mix

Solution

Related Formula

Core Logic

Step 1: Complete Quadratics & Integrate

Step 2: Collect Like Terms

Step 3: Extract Coefficients

Pattern Recognition

Chapter Mix

Solution

Related Formula

Core Logic

Step 1: Factoring out common algebraic terms

Pattern Recognition

Chapter Mix

Solution

Related Formula

Core Logic

Step 1: Solve the First-Order Differential Equation

Step 2: Apply the Boundary Condition

Step 3: Evaluate at target point x = 2

Pattern Recognition

Chapter Mix

Solution

Related Formula

Core Logic

Step 1: Simplify the Integral

Step 2: Solve for Constant C and Find f(1)

Pattern Recognition

Chapter Mix

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$6\int₀π|( 3x+ 2x+ x)|dx$ is equal to.....