JEE Main 2023 (12 April Shift 1)

LEVELJEE Main

Animated Solution for Mathematics - Indefinite Integration: Let $I (x) = \int \frac{x + 7}{x} d x$ and $I (9) = 12 + 7 lo g_{e} 7$ . If $I (1) = α + 7 lo g_{e} (1 + 22)$ , then $α^{4}$ is equal to _____.

Q: Wait, why do we choose $t^2$ instead of just $t$?

Great question! If you set $t = \sqrt{\frac{x+7}{x}}$, you would have to square it anyway to isolate $x$. By setting $t^2 = \frac{x+7}{x}$ from the start, we avoid dealing with square roots in our algebraic manipulations. It keeps the algebra clean and prevents errors.

Enter Numerical Value:

Progressive Hints

Hint 1

The Spark: Look at the structure

\frac{x + 7}{x}

. If you try to integrate this directly, you will get stuck. What if you define the entire radical as a new variable

t

? Let

t = \frac{x + 7}{x}

. This transforms the radical into a simple algebraic term, setting the stage for a much cleaner integration.

Hint 2

The Arsenal: Once you have substituted, you will find yourself with an integral of a rational function in

t

. Remember your toolkit: Integration by Parts is your best friend here. Specifically, look for a way to split the integrand into

u

and

d v

such that the resulting integral is simpler. The choice of

d v

is critical—aim for a term that is easily integrable.

Hint 3

The Execution: Start by setting

t^{2} = 1 + \frac{7}{x}

. This implies

x = \frac{7}{t ^{2} - 1}

. Now, differentiate this to find

d x

in terms of

d t

. Your integral will transform into

I = - 14 \int \frac{t ^{2}}{( t ^{2} - 1 ) ^{2}} d t

. From here, use Integration by Parts with

u = t

and

d v = \frac{t}{( t ^{2} - 1 ) ^{2}} d t

Visualized Solution

Introduction to the Integral $I (x)$

Given integral: $I (x) = \int \frac{x + 7}{x} d x$
Boundary condition: $I (9) = 12 + 7 lo g_{e} 7$
Target: Find $α^{4}$ where $I (1) = α + 7 lo g_{e} (1 + 22)$

Choosing the Substitution

Let $\frac{x + 7}{x} = t^{2}$
This implies $1 + \frac{7}{x} = t^{2}$
Rearranging for $x$ : $\frac{7}{x} = t^{2} - 1$

Expressing $x$ in terms of $t$

$x = \frac{7}{t ^{2} - 1}$
We will use this to find $d x$ in terms of $d t$ .

Calculating the Differential $d x$

Differentiating $x$ with respect to $t$ :
$\frac{d x}{d t} = 7 \cdot \frac{d}{d t} (t^{2} - 1)^{- 1}$
$\frac{d x}{d t} = 7 \cdot (- 1) (t^{2} - 1)^{- 2} \cdot (2 t)$
$d x = \frac{- 14 t}{( t ^{2} - 1 ) ^{2}} d t$

Transforming the Integral

Substitute back into $I (x) = \int \frac{x + 7}{x} d x$ :
$I = \int t \cdot (\frac{- 14 t}{( t ^{2} - 1 ) ^{2}}) d t$
$I = - 14 \int \frac{t ^{2}}{( t ^{2} - 1 ) ^{2}} d t$

Integration by Parts Setup

Using Integration by Parts: $\int u d v = uv - \int v d u$
Let $u = t ⟹ d u = d t$
Let $d v = \frac{t}{( t ^{2} - 1 ) ^{2}} d t ⟹ v = - \frac{1}{2 ( t ^{2} - 1 )}$

Executing Integration by Parts

$I = - 14 [- \frac{t}{2 ( t ^{2} - 1 )} - \int - \frac{1}{2 ( t ^{2} - 1 )} d t]$
$I = \frac{7 t}{t ^{2} - 1} - 7 \int \frac{1}{t ^{2} - 1} d t$
Using $\int \frac{1}{t ^{2} - a ^{2}} d t = \frac{1}{2 a} ln \frac{t - a}{t + a}$ :
$I = \frac{7 t}{t ^{2} - 1} - \frac{7}{2} ln \frac{t - 1}{t + 1} + C$

Back Substitution to $x$

Substitute $t^{2} - 1 = \frac{7}{x}$ and $t = \frac{x + 7}{x}$ :
Term 1: $\frac{7 t}{t ^{2} - 1} = \frac{7 t}{\frac{7}{x}} = x t = x \frac{x + 7}{x} = x (x + 7)$
Term 2: $- \frac{7}{2} ln \frac{t - 1}{t + 1} = \frac{7}{2} ln \frac{t + 1}{t - 1}$
$I (x) = x^{2} + 7 x + \frac{7}{2} ln \frac{t + 1}{t - 1} + C$

Simplifying the Logarithmic Term

$\frac{t + 1}{t - 1} = \frac{( t + 1 ) ^{2}}{t ^{2} - 1} = \frac{( \frac{x + 7}{x} + 1 ) ^{2}}{\frac{7}{x}} = \frac{( x + 7 + x ) ^{2}}{7}$
$I (x) = x^{2} + 7 x + \frac{7}{2} ln (\frac{( x + 7 + x ) ^{2}}{7}) + C$
$I (x) = x^{2} + 7 x + 7 ln (x + 7 + x) - \frac{7}{2} ln 7 + C$

Finding the Constant $C$

Given $I (9) = 12 + 7 lo g_{e} 7$
$I (9) = 81 + 63 + 7 ln (16 + 9) - \frac{7}{2} ln 7 + C$
$12 + 7 ln 7 = 12 + 7 ln 7 - \frac{7}{2} ln 7 + C$
$C = \frac{7}{2} ln 7$

Calculating $I (1)$ and $α$

$I (1) = 1^{2} + 7 (1) + 7 ln (8 + 1) - \frac{7}{2} ln 7 + \frac{7}{2} ln 7$
$I (1) = 8 + 7 ln (22 + 1)$
$I (1) = 22 + 7 ln (1 + 22)$
Comparing with $I (1) = α + 7 lo g_{e} (1 + 22)$ :
$α = 22$

Final Result: $α^{4}$

$α = 22$
$α^{4} = (22)^{4}$
$α^{4} = 2^{4} \cdot (2)^{4}$
$α^{4} = 16 \cdot 4 = 64$
Final Answer: 64

00:00 / 00:00

The Sigma Insight: Integration by Substitution

Analyzing the Setup

Welcome, future engineer. Today, we are going to dissect a problem that looks intimidating at first glance but unfolds with the grace of a well-choreographed dance. We are dealing with the integral:

I (x) = \int \frac{x + 7}{x} d x

The presence of the square root and the variable in the denominator suggests that a direct approach will lead to a dead end. We need a transformation. By setting

t = \frac{x + 7}{x}

, we are essentially mapping our complex variable

x

into a new domain

t

where the radical vanishes. This is the heart of the substitution method: finding a coordinate system where the problem becomes simple.

The Substitution Strategy

When we set

t^{2} = 1 + \frac{7}{x}

, we are not just manipulating symbols; we are simplifying the geometry of the integrand. This implies:

x = \frac{7}{t ^{2} - 1}

Now, we need to find

d x

. Differentiating this with respect to

t

gives us:

d x = \frac{- 14 t}{( t ^{2} - 1 ) ^{2}} d t

This is a crucial step. If you miss that negative sign, the entire calculation will drift off course. Now, look at the integral: it transforms into:

I = - 14 \int \frac{t ^{2}}{( t ^{2} - 1 ) ^{2}} d t

This is where the magic of Integration by Parts happens.

The Integration by Parts

This new integral looks intimidating, but do not be afraid. We split the integrand into

u = t

and

d v = \frac{t}{( t ^{2} - 1 ) ^{2}} d t

. This choice is deliberate. It reduces the power of the denominator, leading us to a standard integral form.

After performing the integration and substituting back, we arrive at:

I (x) = x^{2} + 7 x + 7 ln (x + 7 + x) - \frac{7}{2} ln 7 + C

The boundary condition

I (9) = 12 + 7 ln 7

is our key to finding

C

. When we plug in

x = 9

, the constant

C

emerges as

\frac{7}{2} ln 7

The Final Triumph

Finally, calculating

I (1)

becomes a simple matter of substitution. The terms align, the logarithms simplify, and we find

α = 22

Squaring this twice to get

α^{4}

yields 64.

It is a beautiful journey from complexity to clarity. Remember, in JEE Advanced, the complexity of the problem is often just a veil; once you find the right substitution, the path to the solution becomes clear and inevitable.

Common Traps & FAQs

Wait, why do we choose

t^{2}

instead of just

t

Great question! If you set

t = \frac{x + 7}{x}

, you would have to square it anyway to isolate

x

. By setting

t^{2} = \frac{x + 7}{x}

from the start, we avoid dealing with square roots in our algebraic manipulations. It keeps the algebra clean and prevents errors.

Why does the constant

C

seem to disappear?

It does not disappear! It is just that the boundary condition

I (9)

and the target

I (1)

both contain the same constant terms. When we solve for

C

using

I (9)

, we find its exact value. When we plug that

C

back into

I (1)

, the terms involving

ln 7

cancel out, leaving us with a clean value for

α

Is there a faster way than Integration by Parts?

While you could try partial fractions, Integration by Parts is actually quite elegant here. It handles the

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

denominator efficiently. Always look for the path that minimizes algebraic complexity, and here, IBP is the most direct route.

Animated Solution for Mathematics - Indefinite Integration: Let $I (x) = \int \frac{x + 7}{x} d x$ and $I (9) = 12 + 7 lo g_{e} 7$ . If $I (1) = α + 7 lo g_{e} (1 + 22)$ , then $α^{4}$ is equal to _____.

Enter Numerical Value:

Visualized Solution

Introduction to the Integral $I (x)$

Choosing the Substitution

Expressing $x$ in terms of $t$

Calculating the Differential $d x$

Transforming the Integral

Integration by Parts Setup

Executing Integration by Parts

Back Substitution to $x$

Simplifying the Logarithmic Term

Finding the Constant $C$

Calculating $I (1)$ and $α$

Final Result: $α^{4}$

The Sigma Insight: Integration by Substitution

Analyzing the Setup

The Substitution Strategy

The Integration by Parts

The Final Triumph

Similar Questions

If $α = \int_{0}^{1} (e^{9 x + 3 t a n^{- 1} x}) (\frac{12 + 9 x ^{2}}{1 + x ^{2}}) d x$ where $tan^{- 1} x$ takes only principal values, then the value of $(lo g_{e} ∣1 + α ∣ - \frac{3 π}{4})$ is

Let $f (x) = \int \frac{d x}{( 3 + 4 x ^{2} ) 4 - 3 x ^{2}}, ∣ x ∣ < \frac{2}{3}$ . If $f (0) = 0$ and $f (1) = \frac{1}{α β} tan^{- 1} (\frac{α}{β}), α, β > 0$ , then $α^{2} + β^{2}$ is equal to _______.

If $\int sec 2 x - 1 d x = α lo g_{e} cos 2 x + β + cos 2 x (1 + cos \frac{1}{β} x) + constant$ , then $β - α$ is equal to ______.

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

Let $I (x) = \int \frac{x + 1}{x ( 1 + x e ^{x} ) ^{2}} d x, x > 0$ . If $lim_{x \to \infty} I (x) = 0$ then $I (1)$ is equal to

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

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

If $\int \frac{1}{5 ( x - 1 ) ^{4} ( x + 3 ) ^{6}} d x = A (\frac{α x - 1}{β x + 3})^{B} + C$ , where $C$ is the constant of integration, then the value of $α + β + 20 A B$ is __________

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

Let $f (x) = \int \frac{7 x ^{10} + 9 x ^{8}}{( 1 + x ^{2} + 2 x ^{9} ) ^{2}} d x, x > 0, lim_{x \to 0} f (x) = 0$ and $f (1) = \frac{1}{4}$ . If $A = 0 \frac{1}{4} α^{2} 0 f^{'} (1) 4 111$ and $B = adj (adj A)$ be such that $∣ B ∣ = 81$ , then $α^{2}$ is equal to

Animated Solution for Mathematics - Indefinite Integration: .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Let I(x)=∫xx+7​​dx and I(9)=12+7loge​7. If I(1)=α+7loge​(1+22​), then α4 is equal to _____.

Enter Numerical Value:

Visualized Solution

Introduction to the Integral .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } I(x)

Choosing the Substitution

Expressing .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

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

Integration by Parts Setup

Executing Integration by Parts

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

Simplifying the Logarithmic Term

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

Final Result: .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } α4

The Sigma Insight: Integration by Substitution

Analyzing the Setup

The Substitution Strategy

The Integration by Parts

The Final Triumph

Similar Questions

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } If α=∫01​(e9x+3tan−1x)(1+x212+9x2​)dx where tan−1x takes only principal values, then the value of (loge​∣1+α∣−43π​) is

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Let f(x)=∫(3+4x2)4−3x2​dx​,∣x∣<3​2​. If f(0)=0 and f(1)=αβ1​tan−1(βα​),α,β>0, then α2+β2 is equal to _______.

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } If ∫sec2x−1​dx=αloge​​cos2x+β+cos2x(1+cosβ1​x)​​+constant, then β−α is equal to ______.

.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:

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Let I(x)=∫x(1+xex)2x+1​dx,x>0. If limx→∞​I(x)=0 then I(1) is equal to

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } If f(x)=∫(x2+1+2x7)25x8+7x6​dx,(x≥0),f(0)=0 and f(1)=K1​, then the value of K is

.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

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } If ∫5(x−1)4(x+3)6​1​dx=A(βx+3αx−1​)B+C, where C is the constant of integration, then the value of α+β+20AB is __________

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Let f(x)=∫(x2+1)(x2+3)2x​dx. If f(3)=21​(loge​5−loge​6), then f(4) is equal to

Animated Solution for Mathematics - Indefinite Integration: Let $I (x) = \int \frac{x + 7}{x} d x$ and $I (9) = 12 + 7 lo g_{e} 7$ . If $I (1) = α + 7 lo g_{e} (1 + 22)$ , then $α^{4}$ is equal to _____.

Introduction to the Integral $I (x)$

Expressing $x$ in terms of $t$

Calculating the Differential $d x$

Back Substitution to $x$

Finding the Constant $C$

Calculating $I (1)$ and $α$

Final Result: $α^{4}$

If $α = \int_{0}^{1} (e^{9 x + 3 t a n^{- 1} x}) (\frac{12 + 9 x ^{2}}{1 + x ^{2}}) d x$ where $tan^{- 1} x$ takes only principal values, then the value of $(lo g_{e} ∣1 + α ∣ - \frac{3 π}{4})$ is

Let $f (x) = \int \frac{d x}{( 3 + 4 x ^{2} ) 4 - 3 x ^{2}}, ∣ x ∣ < \frac{2}{3}$ . If $f (0) = 0$ and $f (1) = \frac{1}{α β} tan^{- 1} (\frac{α}{β}), α, β > 0$ , then $α^{2} + β^{2}$ is equal to _______.

If $\int sec 2 x - 1 d x = α lo g_{e} cos 2 x + β + cos 2 x (1 + cos \frac{1}{β} x) + constant$ , then $β - α$ is equal to ______.

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

Let $I (x) = \int \frac{x + 1}{x ( 1 + x e ^{x} ) ^{2}} d x, x > 0$ . If $lim_{x \to \infty} I (x) = 0$ then $I (1)$ is equal to

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

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

If $\int \frac{1}{5 ( x - 1 ) ^{4} ( x + 3 ) ^{6}} d x = A (\frac{α x - 1}{β x + 3})^{B} + C$ , where $C$ is the constant of integration, then the value of $α + β + 20 A B$ is __________

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