Evaluate ∫ x² dx / (a + bx)²

MathematicsIntegration by SubstitutionJEE Advanced 1979Easy

Evaluate $\int \frac{x ^{2} d x}{( a + b x ) ^{2}}$

Answer:

\frac{1}{b ^{3}} [a + b x - 2 a lo g ∣ a + b x ∣ - \frac{a ^{2}}{a + b x}] + C

Visualized Solution (English)

Problem Analysis and Strategy

Given integral: $I = \int \frac{x ^{2}}{( a + b x ) ^{2}} d x$
The denominator contains a linear expression $(a + b x)$ raised to a power.
Strategy: Use substitution to simplify the denominator into a single variable.

Defining the Substitution

Let $a + b x = t$
Differentiating both sides with respect to $x$ :
$b d x = d t \Rightarrow d x = \frac{d t}{b}$

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

From $a + b x = t$ , we can isolate $x$ :
$b x = t - a$
$x = \frac{t - a}{b}$

Substituting into the Integral

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

Simplifying Constants

Pull out the constant terms:
$I = \frac{1}{b ^{2} \cdot b} \int \frac{( t - a ) ^{2}}{t ^{2}} d t$
$I = \frac{1}{b ^{3}} \int \frac{( t - a ) ^{2}}{t ^{2}} d t$

Expanding the Numerator

Expand $(t - a)^{2}$ using the identity $(u - v)^{2} = u^{2} - 2 uv + v^{2}$ :
$I = \frac{1}{b ^{3}} \int \frac{t ^{2} - 2 a t + a ^{2}}{t ^{2}} d t$

Splitting the Fraction

Divide each term in the numerator by $t^{2}$ :
$I = \frac{1}{b ^{3}} \int (\frac{t ^{2}}{t ^{2}} - \frac{2 a t}{t ^{2}} + \frac{a ^{2}}{t ^{2}}) d t$
$I = \frac{1}{b ^{3}} \int (1 - \frac{2 a}{t} + \frac{a ^{2}}{t ^{2}}) d t$

Integrating Term by Term

Integrate each term with respect to $t$ :
$\int 1 d t = t$
$\int \frac{2 a}{t} d t = 2 a lo g ∣ t ∣$
$\int \frac{a ^{2}}{t ^{2}} d t = a^{2} \int t^{- 2} d t = a^{2} (\frac{t ^{- 1}}{- 1}) = - \frac{a ^{2}}{t}$
Combining them: $I = \frac{1}{b ^{3}} [t - 2 a lo g ∣ t ∣ - \frac{a ^{2}}{t}] + C$

Back Substitution

Substitute $t = a + b x$ back into the expression:
$I = \frac{1}{b ^{3}} [(a + b x) - 2 a lo g ∣ a + b x ∣ - \frac{a ^{2}}{a + b x}] + C$

Final Result and Key Takeaways

Final Answer: $I = \frac{1}{b ^{3}} [a + b x - 2 a lo g ∣ a + b x ∣ - \frac{a ^{2}}{a + b x}] + C$
Key Takeaway: Substitution $t = a + b x$ simplified the denominator, allowing for term-by-term integration.
Challenge: Try evaluating $\int \frac{x ^{3}}{( a + b x ) ^{2}} d x$ using the same method.

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Conceptually Similar Problems

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

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lo g (\frac{1 + x e ^{x}}{x e ^{x}}) - \frac{1}{1 + x e ^{x}} + C

MathematicsFundamental Theorem & Properties of Definite IntegralsJEE Advanced 1993Moderate

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Evaluate $\int_{2}^{3} \frac{2 x ^{5} + x ^{4} - 2 x ^{3} + 2 x ^{2} + 1}{( x ^{2} + 1 ) ( x ^{4} - 1 )} d x$

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\frac{1}{2} lo g 6 - \frac{1}{10}

MathematicsIntegration by SubstitutionJEE Advanced 1984Easy

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- (1 + \frac{1}{x ^{4}})^{1/4} + C

MathematicsFundamental Theorem & Properties of Definite IntegralsJEE Advanced 1988Moderate

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Evaluate $\int_{0}^{1} lo g [1 - x + 1 + x] d x$

Answer:

\frac{1}{2} [lo g 2 + \frac{π}{2} - 1]

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Evaluate $\int \frac{( c o s 2 x ) ^{1/2}}{s i n x} d x$

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\frac{1}{2} lo g \frac{2 + 1 - t a n ^{2} x}{2 - 1 - t a n ^{2} x} - lo g (cot x + cot^{2} x - 1) + C

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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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\frac{1}{2} lo g ∣ sin x - cos x ∣ + \frac{x}{2} + C

MathematicsIntegration using Partial FractionsJEE Advanced 1999Moderate

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Integrate $\int \frac{x ^{3} + 3 x + 2}{( x ^{2} + 1 ) ^{2} ( x + 1 )} d x$

Answer:

- \frac{1}{2} lo g ∣ x + 1∣ + \frac{1}{4} lo g (x^{2} + 1) + \frac{3}{2} tan^{- 1} x + \frac{x}{1 + x ^{2}} + C

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Evaluate the following $\int \frac{1 - x}{1 + x} d x$

Answer:

- 2 1 - x + cos^{- 1} x + x 1 - x + C

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Evaluate $\int_{0}^{π} \frac{x s i n 2 x s i n ( \frac{π}{2} c o s x )}{2 x - π} d x$

Answer:

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MathematicsEvaluation of Special Integral FormsJEE Main 2005Easy

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$\int {\frac{l o g x - 1}{1 + ( l o g x ) ^{2}}}^{2} d x$ is equal to

(A)

\frac{l o g x}{( l o g x ) ^{2} + 1} + C

(B)

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

(C)

\frac{x e ^{x}}{1 + x ^{2}} + C

(D)

\frac{x}{( l o g x ) ^{2} + 1} + C

Answer:D

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Evaluate ∫(a+bx)2x2dx​

Visualized Solution (English)

Problem Analysis and Strategy

Defining 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

Substituting into the Integral

Simplifying Constants

Expanding the Numerator

Splitting the Fraction

Integrating Term by Term

Back Substitution

Final Result and Key Takeaways

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