JEE Main 2024 (27 Jan Shift 2)

LEVELJEE Main

Animated Solution for Mathematics - Indefinite Integration: The integral $\int \frac{( x ^{8} - x ^{2} ) d x}{( x ^{12} + 3 x ^{6} + 1 ) t a n ^{- 1} ( x ^{3} + \frac{1}{x ^{3}} )}$ equal to :

Q: I got stuck at the derivative of $x^3 + \frac{1}{x^3}$. How do I simplify that to match the numerator?

Don't worry, it's just algebra! The derivative is $3x^2 - 3x^{-4}$. If you factor out the $3$, you get $3(x^2 - \frac{1}{x^4})$. To combine these, find a common denominator: $3(\frac{x^6 - 1}{x^4})$. When you multiply this by the $\frac{1}{1+u^2}$ term, the $x^6$ in the denominator of that term will interact with the $x^4$ in the denominator of your derivative, leaving you with the $x^2$ needed to create the $x^8 - x^2$ in the numerator. It's a beautiful cancellation!

Q: Why does the $x^6$ term appear in the numerator after differentiation?

That comes from the $\frac{1}{1+u^2}$ part of the chain rule. When you expand $1 + (x^3 + \frac{1}{x^3})^2$, you get $1 + x^6 + 2 + \frac{1}{x^6}$. This simplifies to $x^6 + 3 + \frac{1}{x^6}$. When you write this as a single fraction, it becomes $\frac{x^{12} + 3x^6 + 1}{x^6}$. Since this is in the denominator of the derivative, the $x^6$ flips to the numerator. It's not magic; it's just the result of careful algebraic manipulation!

Select Answer:

Visualized Solution

Analyze the Integrand

$I = \int \frac{x ^{8} - x ^{2}}{( x ^{12} + 3 x ^{6} + 1 ) t a n ^{- 1} ( x ^{3} + \frac{1}{x ^{3}} )} d x$
Identify the complex part of the integrand for potential substitution.

Define Substitution $t$

Let $t = tan^{- 1} (x^{3} + \frac{1}{x ^{3}})$
Goal: Find $d t$ in terms of $d x$ .

Differentiate using Chain Rule

$\frac{d t}{d x} = \frac{1}{1 + ( x ^{3} + \frac{1}{x ^{3}} ) ^{2}} \cdot \frac{d}{d x} (x^{3} + \frac{1}{x ^{3}})$
Apply Chain Rule: $\frac{d}{d x} [tan^{- 1} (u)] = \frac{1}{1 + u ^{2}} \cdot \frac{d u}{d x}$

Compute Inner Derivative

$\frac{d}{d x} (x^{3} + \frac{1}{x ^{3}}) = 3 x^{2} - \frac{3}{x ^{4}}$
$= 3 (x^{2} - \frac{1}{x ^{4}}) = \frac{3 ( x ^{6} - 1 )}{x ^{4}}$

Expand the Denominator

$1 + (x^{3} + \frac{1}{x ^{3}})^{2} = 1 + (x^{6} + \frac{1}{x ^{6}} + 2)$
$= x^{6} + 3 + \frac{1}{x ^{6}}$

Simplify the Fraction

$x^{6} + 3 + \frac{1}{x ^{6}} = \frac{x ^{12} + 3 x ^{6} + 1}{x ^{6}}$
The term becomes $\frac{x ^{6}}{x ^{12} + 3 x ^{6} + 1}$ when inverted.

Combine All Terms for $d t$

$\frac{d t}{d x} = \frac{x ^{6}}{x ^{12} + 3 x ^{6} + 1} \cdot \frac{3 ( x ^{6} - 1 )}{x ^{4}}$
$= \frac{3 x ^{2} ( x ^{6} - 1 )}{x ^{12} + 3 x ^{6} + 1} = \frac{3 ( x ^{8} - x ^{2} )}{x ^{12} + 3 x ^{6} + 1}$

Final Substitution into Integral

$\frac{1}{3} d t = \frac{x ^{8} - x ^{2}}{x ^{12} + 3 x ^{6} + 1} d x$
$I = \int \frac{\frac{1}{3} d t}{t} = \frac{1}{3} \int \frac{d t}{t}$

Integrate with respect to $t$

$I = \frac{1}{3} lo g_{e} ∣ t ∣ + C$
Using the standard integral $\int \frac{1}{x} d x = lo g_{e} ∣ x ∣ + C$ .

Back-Substitution and Final Form

$I = \frac{1}{3} lo g_{e} ∣ tan^{- 1} (x^{3} + \frac{1}{x ^{3}}) ∣ + C$
$I = lo g_{e} (tan^{- 1} (x^{3} + \frac{1}{x ^{3}}))^{\frac{1}{3}} + C$

Summary and Conclusion

Key Takeaway: Always look for the derivative of complex terms like $tan^{- 1} (f (x))$ within the integrand.
Next Challenge: Try solving $\int \frac{( x ^{4} - 1 ) d x}{( x ^{6} + x ^{4} + x ^{2} + 1 ) t a n ^{- 1} ( \frac{x ^{2} + 1}{x} )}$ .

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The Sigma Insight: Integration by Substitution

Analyzing the Setup

The given integral is:

I = \int \frac{( x ^{8} - x ^{2} ) d x}{( x ^{12} + 3 x ^{6} + 1 ) tan ^{- 1} ( x ^{3} + \frac{1}{x ^{3}} )}

The denominator contains a complex combination of high-degree polynomials and an inverse trigonometric function. In JEE Advanced problems, such complexity often masks a hidden symmetry that can be unlocked through strategic substitution.

The Art of Substitution

We focus on the term

t = tan^{- 1} (x^{3} + \frac{1}{x ^{3}})

. Inverse trigonometric functions are often the "beacons" of an integral, as their derivatives frequently simplify into algebraic expressions that match the numerator.

Let

u = x^{3} + \frac{1}{x ^{3}}

. Then

t = tan^{- 1} (u)

. Differentiating with respect to

x

using the chain rule:

\frac{d t}{d x} = \frac{1}{1 + u ^{2}} \cdot \frac{d u}{d x}

Evaluating the Differential

First, we calculate the derivative of

u

\frac{d u}{d x} = 3 x^{2} - 3 x^{- 4} = 3 (x^{2} - \frac{1}{x ^{4}}) = \frac{3 ( x ^{6} - 1 )}{x ^{4}}

Next, we simplify the term

1 + u^{2}

1 + u^{2} = 1 + (x^{3} + \frac{1}{x ^{3}})^{2} = 1 + x^{6} + 2 + x^{- 6} = \frac{x ^{12} + 3 x ^{6} + 1}{x ^{6}}

Combining these into the expression for

d t

d t = (\frac{x ^{6}}{x ^{12} + 3 x ^{6} + 1}) \cdot (\frac{3 ( x ^{6} - 1 )}{x ^{4}}) d x = \frac{3 ( x ^{8} - x ^{2} )}{x ^{12} + 3 x ^{6} + 1} d x

Final Calculation

We observe that the expression

\frac{( x ^{8} - x ^{2} ) d x}{x ^{12} + 3 x ^{6} + 1}

is exactly equal to

\frac{1}{3} d t

. The integral now simplifies beautifully:

I = \frac{1}{3} \int \frac{d t}{t}

Integrating this yields:

I = \frac{1}{3} ln ∣ t ∣ + C

Substituting back for

t

, we arrive at the final result:

$I = \frac{1}{3} ln tan^{- 1} (x^{3} + \frac{1}{x ^{3}}) + C$

Common Traps & FAQs

Why did you choose

t = tan^{- 1} (x^{3} + \frac{1}{x ^{3}})

? How do I know what to substitute in such complex integrals?

Great question! In JEE Advanced, when you see a function like

tan^{- 1} (f (x))

ln (g (x))

in an integral, it is almost always the key. These functions have derivatives that are algebraic, and the problem is designed so that the rest of the integrand (the 'algebraic mess') is actually the derivative of that function. Always look for the most 'transcendental' or complex-looking part of the integrand and test its derivative.

I got stuck at the derivative of

x^{3} + \frac{1}{x ^{3}}

. How do I simplify that to match the numerator?

Don't worry, it's just algebra! The derivative is

3 x^{2} - 3 x^{- 4}

. If you factor out the

3

, you get

3 (x^{2} - \frac{1}{x ^{4}})

. To combine these, find a common denominator:

3 (\frac{x ^{6} - 1}{x ^{4}})

. When you multiply this by the

\frac{1}{1 + u ^{2}}

term, the

x^{6}

in the denominator of that term will interact with the

x^{4}

in the denominator of your derivative, leaving you with the

x^{2}

needed to create the

x^{8} - x^{2}

in the numerator. It's a beautiful cancellation!

Why does the

x^{6}

term appear in the numerator after differentiation?

That comes from the

\frac{1}{1 + u ^{2}}

part of the chain rule. When you expand

1 + (x^{3} + \frac{1}{x ^{3}})^{2}

, you get

1 + x^{6} + 2 + \frac{1}{x ^{6}}

. This simplifies to

x^{6} + 3 + \frac{1}{x ^{6}}

. When you write this as a single fraction, it becomes

\frac{x ^{12} + 3 x ^{6} + 1}{x ^{6}}

. Since this is in the denominator of the derivative, the

x^{6}

flips to the numerator. It's not magic; it's just the result of careful algebraic manipulation!

Animated Solution for Mathematics - Indefinite Integration: The integral $\int \frac{( x ^{8} - x ^{2} ) d x}{( x ^{12} + 3 x ^{6} + 1 ) t a n ^{- 1} ( x ^{3} + \frac{1}{x ^{3}} )}$ equal to :

Select Answer:

Visualized Solution

Analyze the Integrand

Define Substitution $t$

Differentiate using Chain Rule

Compute Inner Derivative

Expand the Denominator

Simplify the Fraction

Combine All Terms for $d t$

Final Substitution into Integral

Integrate with respect to $t$

Back-Substitution and Final Form

Summary and Conclusion

The Sigma Insight: Integration by Substitution

Analyzing the Setup

The Art of Substitution

Evaluating the Differential

Final Calculation

Similar Questions

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

The integral $\int \frac{2 x ^{12} + 5 x ^{9}}{( x ^{5} + x ^{3} + 1 ) ^{3}} d x$ is equal to

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

Find the indefinite integral $\int (\frac{1}{3 x + 4 x} + \frac{l n ( 1 + 6 x )}{3 x + x}) d x$

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

Evaluate $\int sin^{- 1} (\frac{2 x + 2}{4 x ^{2} + 8 x + 13}) d x$

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

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

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 :

Find integration $\int \frac{d x}{( x - 3 ) ^{6/7} ( x + 4 ) ^{8/7}}$

Animated Solution for Mathematics - Indefinite Integration: .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } The integral ∫(x12+3x6+1)tan−1(x3+x31​)(x8−x2)dx​ equal to :

Select Answer:

Visualized Solution

Analyze the Integrand

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

Differentiate using Chain Rule

Compute Inner Derivative

Expand the Denominator

Simplify the Fraction

Combine All Terms for .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } dt

Final Substitution into Integral

Integrate with respect to .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } t

Back-Substitution and Final Form

Summary and Conclusion

The Sigma Insight: Integration by Substitution

Analyzing the Setup

The Art of Substitution

Evaluating the Differential

Final Calculation

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Animated Solution for Mathematics - Indefinite Integration: The integral $\int \frac{( x ^{8} - x ^{2} ) d x}{( x ^{12} + 3 x ^{6} + 1 ) t a n ^{- 1} ( x ^{3} + \frac{1}{x ^{3}} )}$ equal to :

Define Substitution $t$

Combine All Terms for $d t$

Integrate with respect to $t$

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

The integral $\int \frac{2 x ^{12} + 5 x ^{9}}{( x ^{5} + x ^{3} + 1 ) ^{3}} d x$ is equal to

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

Find the indefinite integral $\int (\frac{1}{3 x + 4 x} + \frac{l n ( 1 + 6 x )}{3 x + x}) d x$

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

Evaluate $\int sin^{- 1} (\frac{2 x + 2}{4 x ^{2} + 8 x + 13}) d x$

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

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

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 :

Find integration $\int \frac{d x}{( x - 3 ) ^{6/7} ( x + 4 ) ^{8/7}}$