$\int \left\{ \frac{\log x - 1}{1 + (\log x)^2} \right\}^2 dx$ is equal to Options: (A) $\frac{\log x}{(\log x)^2 + 1} + C$ (B) $\frac{x}{x^2 + 1} + C$ (C) $\frac{xe^x}{1 + x^2} + C$ (D) $\frac{x}{(\log x)^2 + 1} + C$

### Analyzing the Setup The integral we are tasked to solve is: $$I = \int \left\{ \frac{\log x - 1}{1 + (\log x)^2} \right\}^2 dx$$ To simplify this expression, we perform the substitution $t = \log x$. This implies $x = e^t$, and consequently, the differential becomes $dx = e^t dt$. ### Transforming the Integral Substituting these values into the original integral, we obtain: $$I = \int \left( \frac{t - 1}{1 + t^2} \right)^2 e^t dt$$ Expanding the numerator, we have $(t - 1)^2 = t^2 - 2t + 1$. We can rewrite this as $(t^2 + 1) - 2t$, allowing us to express the integral as: $$I = \int e^t \left[ \frac{t^2 + 1 - 2t}{(1 + t^2)^2} \right] dt$$ ### Applying Algebraic Surgery We now split the fraction into two distinct parts to reveal the underlying structure: $$I = \int e^t \left[ \frac{t^2 + 1}{(1 + t^2)^2} - \frac{2t}{(1 + t^2)^2} \right] dt$$ Simplifying the first term by canceling the common factor $(1 + t^2)$, we get: $$I = \int e^t \left[ \frac{1}{1 + t^2} - \frac{2t}{(1 + t^2)^2} \right] dt$$ ### Identifying the Pattern This expression matches the standard form $\int e^t [f(t) + f'(t)] dt$. Let us define $f(t) = \frac{1}{1 + t^2}$. Calculating the derivative $f'(t)$ using the chain rule: $$f'(t) = \frac{d}{dt} (1 + t^2)^{-1} = -1(1 + t^2)^{-2} \cdot (2t) = -\frac{2t}{(1 + t^2)^2}$$ Since our integral is exactly in the form $\int e^t [f(t) + f'(t)] dt$, we know the solution is $e^t f(t) + C$. ### Final Calculation Substituting $f(t)$ back into the result, we have: $$I = e^t \cdot \frac{1}{1 + t^2} + C$$ Finally, replacing $t$ with $\log x$ and $e^t$ with $x$, we arrive at the final answer: **$$I = \frac{x}{1 + (\log x)^2} + C$$**

JEE Main 2005

LEVELJEE Main

Animated Solution for Mathematics - Indefinite Integration: $\int {\frac{l o g x - 1}{1 + ( l o g x ) ^{2}}}^{2} d x$ is equal to

Q: Is there a way to solve this without the $e^t [f(t) + f'(t)]$ form?

While there might be other methods, the $e^t [f(t) + f'(t)]$ form is the most elegant and efficient way to solve this. It is a fundamental tool in your JEE arsenal. Trying to solve it without this form would likely involve much more tedious integration by parts, which increases the chance of errors. Always look for this pattern when you see $e^t$ in an integral!

Select Answer:

Visualized Solution

Analyzing the Integral

Given Integral: $I = \int {\frac{l o g x - 1}{1 + ( l o g x ) ^{2}}}^{2} d x$
Objective: Simplify the complex logarithmic integrand.

Applying Substitution: $t = lo g x$

Let $lo g x = t$
This implies $x = e^{t}$

Finding $d x$

Differentiating $x = e^{t}$ with respect to $t$ :
$\frac{d x}{d t} = e^{t} ⟹ d x = e^{t} d t$

Rewriting the Integral

Substitute $t$ and $d x$ into the integral:
$I = \int (\frac{t - 1}{1 + t ^{2}})^{2} e^{t} d t$
$I = \int e^{t} \frac{( t - 1 ) ^{2}}{( 1 + t ^{2} ) ^{2}} d t$

Expanding the Numerator

Expand the numerator using $(a - b)^{2} = a^{2} - 2 ab + b^{2}$ :
$(t - 1)^{2} = t^{2} - 2 t + 1$
Rewrite the integral: $I = \int e^{t} [\frac{t ^{2} + 1 - 2 t}{( 1 + t ^{2} ) ^{2}}] d t$

Splitting the Fraction

Split the integrand into two parts:
$I = \int e^{t} [\frac{t ^{2} + 1}{( 1 + t ^{2} ) ^{2}} - \frac{2 t}{( 1 + t ^{2} ) ^{2}}] d t$
Simplify the first term: $I = \int e^{t} [\frac{1}{1 + t ^{2}} - \frac{2 t}{( 1 + t ^{2} ) ^{2}}] d t$

Identifying the Standard Form

The integral resembles the standard form: $\int e^{t} [f (t) + f^{'} (t)] d t$
Let $f (t) = \frac{1}{1 + t ^{2}}$

Differentiating $f (t)$

Differentiating $f (t)$ with respect to $t$ :
$f^{'} (t) = \frac{d}{d t} (1 + t^{2})^{- 1} = - 1 (1 + t^{2})^{- 2} \cdot (2 t)$
$f^{'} (t) = - \frac{2 t}{( 1 + t ^{2} ) ^{2}}$

Applying the Integration Rule

The integral is exactly $\int e^{t} [f (t) + f^{'} (t)] d t$
Using the property: $\int e^{t} [f (t) + f^{'} (t)] d t = e^{t} f (t) + C$
$I = e^{t} \cdot \frac{1}{1 + t ^{2}} + C$

Final Back-Substitution

Substitute $t = lo g x$ and $e^{t} = x$ back into the result:
$I = \frac{x}{1 + ( l o g x ) ^{2}} + C$
Final Answer: $\frac{x}{( l o g x ) ^{2} + 1} + C$

00:00 / 00:00

The Sigma Insight: Evaluation of Special Integral Forms

Analyzing the Setup

The integral we are tasked to solve is:

I = \int {\frac{lo g x - 1}{1 + ( lo g x ) ^{2}}}^{2} d x

To simplify this expression, we perform the substitution

t = lo g x

. This implies

x = e^{t}

, and consequently, the differential becomes

d x = e^{t} d t

Transforming the Integral

Substituting these values into the original integral, we obtain:

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

Expanding the numerator, we have

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

. We can rewrite this as

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

, allowing us to express the integral as:

I = \int e^{t} [\frac{t ^{2} + 1 - 2 t}{( 1 + t ^{2} ) ^{2}}] d t

Applying Algebraic Surgery

We now split the fraction into two distinct parts to reveal the underlying structure:

I = \int e^{t} [\frac{t ^{2} + 1}{( 1 + t ^{2} ) ^{2}} - \frac{2 t}{( 1 + t ^{2} ) ^{2}}] d t

Simplifying the first term by canceling the common factor

(1 + t^{2})

, we get:

I = \int e^{t} [\frac{1}{1 + t ^{2}} - \frac{2 t}{( 1 + t ^{2} ) ^{2}}] d t

Identifying the Pattern

This expression matches the standard form

\int e^{t} [f (t) + f^{'} (t)] d t

. Let us define

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

Calculating the derivative

f^{'} (t)

using the chain rule:

f^{'} (t) = \frac{d}{d t} (1 + t^{2})^{- 1} = - 1 (1 + t^{2})^{- 2} \cdot (2 t) = - \frac{2 t}{( 1 + t ^{2} ) ^{2}}

Since our integral is exactly in the form

\int e^{t} [f (t) + f^{'} (t)] d t

, we know the solution is

e^{t} f (t) + C

Final Calculation

Substituting

f (t)

back into the result, we have:

I = e^{t} \cdot \frac{1}{1 + t ^{2}} + C

Finally, replacing

t

with

lo g x

and

e^{t}

with

x

, we arrive at the final answer:

$I = \frac{x}{1 + ( lo g x ) ^{2}} + C$

Common Traps & FAQs

Why do we substitute

t = lo g x

instead of something else?

Great question! When you see a complex expression involving

lo g x

and its powers, it is a classic signal that a substitution will simplify the structure. By setting

t = lo g x

, we transform a logarithmic integral into an exponential one, which is often much easier to handle using standard integration techniques.

How do I know when to split the fraction?

Splitting the fraction is a strategic move. When you see a denominator like

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

and a numerator that can be expanded into terms involving

t^{2} + 1

and

t

, it is a strong hint that you can create a form that matches the derivative of a function. Always look for ways to simplify the expression into a sum of a function and its derivative.

Is there a way to solve this without the

e^{t} [f (t) + f^{'} (t)]

form?

While there might be other methods, the

e^{t} [f (t) + f^{'} (t)]

form is the most elegant and efficient way to solve this. It is a fundamental tool in your JEE arsenal. Trying to solve it without this form would likely involve much more tedious integration by parts, which increases the chance of errors. Always look for this pattern when you see

e^{t}

in an integral!

Animated Solution for Mathematics - Indefinite Integration: $\int {\frac{l o g x - 1}{1 + ( l o g x ) ^{2}}}^{2} d x$ is equal to

Select Answer:

Visualized Solution

Analyzing the Integral

Applying Substitution: $t = lo g x$

Finding $d x$

Rewriting the Integral

Expanding the Numerator

Splitting the Fraction

Identifying the Standard Form

Differentiating $f (t)$

Applying the Integration Rule

Final Back-Substitution

The Sigma Insight: Evaluation of Special Integral Forms

Analyzing the Setup

Transforming the Integral

Applying Algebraic Surgery

Identifying the Pattern

Final Calculation

Similar Questions

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Animated Solution for Mathematics - Indefinite Integration: .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } ∫{1+(logx)2logx−1​}2dx is equal to

Select Answer:

Visualized Solution

Analyzing the Integral

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

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

Rewriting the Integral

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Applying the Integration Rule

Final Back-Substitution

The Sigma Insight: Evaluation of Special Integral Forms

Analyzing the Setup

Transforming the Integral

Applying Algebraic Surgery

Identifying the Pattern

Final Calculation

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Animated Solution for Mathematics - Indefinite Integration: $\int {\frac{l o g x - 1}{1 + ( l o g x ) ^{2}}}^{2} d x$ is equal to

Applying Substitution: $t = lo g x$

Finding $d x$

Differentiating $f (t)$

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Evaluate $\int \frac{s i n x}{s i n x - c o s x} d x$

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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

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