If 4x/(x^2 - 1)^2 = A_1/(x - 1) + A_2/(x - 1)^2 + A_3/(x + 1) + A_4/(x + 1)^2 then A_1 + A_2 + A_3 + A_4 =

The correct answer is Option 3. Full mathematical solution steps and hints are accessible in the interactive StemCET workspace.

Maths Q19 – AP-EAMCET 2026

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Home Maths IntegralsQuestion #203

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

then

A_{1} + A_{2} + A_{3} + A_{4} =

Clue 1 of 3: Partial Fraction Decomposition

This problem involves decomposing a rational function into simpler fractions, which is a standard technique used in algebra and calculus.

🧠 Full Solution Path

Step 1: Simplify the Denominator and Set up the General Equation✦ Active

The given expression for partial fraction decomposition is:

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

First, simplify the denominator on the left side: $(x^{2} - 1)^{2} = ((x - 1) (x + 1))^{2} = (x - 1)^{2} (x + 1)^{2}$ . Now, multiply both sides of the equation by $(x - 1)^{2} (x + 1)^{2}$ to eliminate the denominators:

4 x = A_{1} (x - 1) (x + 1)^{2} + A_{2} (x + 1)^{2} + A_{3} (x + 1) (x - 1)^{2} + A_{4} (x - 1)^{2}

This equation is an identity, meaning it holds true for all valid values of $x$ .

💡 Teacher's Secret Hint

Remember that $(x^{2} - 1)^{2}$ is a difference of squares squared, so it decomposes into $(x - 1)^{2} (x + 1)^{2}$ . This is crucial for correctly setting up the partial fractions.

Step 2: Determine

A_{2}

and

A_{4}

by Substitution○ Expand

To find $A_{2}$ and $A_{4}$ , substitute values of $x$ that make other terms zero:

* Set $x = 1$ :

4 (1) = A_{1} (0) + A_{2} (1 + 1)^{2} + A_{3} (0) + A_{4} (0)

$4 = A_{2} (2)^{2} ⟹ 4 = 4 A_{2} ⟹ A_{2} = 1$

* Set $x = - 1$ :

4 (- 1) = A_{1} (0) + A_{2} (0) + A_{3} (0) + A_{4} (- 1 - 1)^{2}

$- 4 = A_{4} (- 2)^{2} ⟹ - 4 = 4 A_{4} ⟹ A_{4} = - 1$

💡 Teacher's Secret Hint

Substituting the roots of the denominators (poles) is usually the quickest way to find the coefficients of the highest power terms in the partial fraction expansion.

Step 3: Determine

A_{1}

and

A_{3}

by Equating Coefficients○ Expand

Expand the right side of the identity and equate coefficients of like powers of $x$ .

4 x = A_{1} (x^{3} + x^{2} - x - 1) + A_{2} (x^{2} + 2 x + 1) + A_{3} (x^{3} - x^{2} - x + 1) + A_{4} (x^{2} - 2 x + 1)

Group terms by powers of $x$ :

4 x = (A_{1} + A_{3}) x^{3} + (A_{1} + A_{2} - A_{3} + A_{4}) x^{2} + (- A_{1} + 2 A_{2} - A_{3} - 2 A_{4}) x + (- A_{1} + A_{2} + A_{3} + A_{4})

Equate coefficients of $x^{3}$ (since there is no $x^{3}$ term on the left, its coefficient is $0$ ):

0 = A_{1} + A_{3} (Equation 1)

Equate coefficients of $x^{2}$ (since there is no $x^{2}$ term on the left, its coefficient is $0$ ):

0 = A_{1} + A_{2} - A_{3} + A_{4}

Substitute the values $A_{2} = 1$ and $A_{4} = - 1$ found in Step 2:

0 = A_{1} + 1 - A_{3} - 1

0 = A_{1} - A_{3} (Equation 2)

From Equation 1, $A_{3} = - A_{1}$ . Substitute this into Equation 2:

0 = A_{1} - (- A_{1}) ⟹ 0 = 2 A_{1} ⟹ A_{1} = 0

Since $A_{3} = - A_{1}$ , then $A_{3} = 0$ .

💡 Teacher's Secret Hint

Remember to equate coefficients for ALL powers of $x$ present in the expanded form on the right, even if they are zero on the left side of the identity.

Step 4: Calculate the Sum

A_{1} + A_{2} + A_{3} + A_{4}

○ Expand

We have determined all the coefficients:

$A_{1} = 0$

$A_{2} = 1$

$A_{3} = 0$

$A_{4} = - 1$

Now, sum them up:

A_{1} + A_{2} + A_{3} + A_{4} = 0 + 1 + 0 + (- 1) = 0

The sum is $0$ , which corresponds to option 3.

💡 Teacher's Secret Hint

Always double-check your arithmetic, especially when adding positive and negative numbers.

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Maths Question 19 – AP-EAMCET 2026

Clue 1 of 3: Partial Fraction Decomposition

🧠 Full Solution Path