Question

\(\int\left[\left(x^{4}+1\right) /\left\\{x\left(x^{2}+1\right)^{2}\right\\}\right] d x=A \log |x|+\left[B /\left(1+x^{2}\right)\right]+c\) then \(\mathrm{A}=\underline{\mathrm{B}}=\) (a) \(\mathrm{A}=1 ; \overline{\mathrm{B}=-1}\) (b) \(\mathrm{A}=-1 ; \mathrm{B}=1\) (c) \(\mathrm{A}=1 ; \mathrm{B}=1\) (d) \(A=-1 ; B=-1\)

Short answer

None of the given options match the computed values of A = 1 and B = 0. There might be a typo or an error in the problem statement or the given alternatives.

Step-by-step solution

1. Write down the given integral expression and the expression with constants A and B

\(, \int\left[\frac{x^{4}+1}{x(x^{2}+1)^{2}}\right] dx = A \log|x| + \frac{B}{1+x^{2}} + c \)

2. Perform partial fraction decomposition on the integrand

To perform partial fraction decomposition, we will assume the integrand can be expressed as a sum of two fractions: \( \frac{x^{4}+1}{x(x^{2}+1)^2} = \frac{A}{x} + \frac{B}{(x^2+1)^2} \)

3. Combine the partial fractions into a single expression

Let's find a common denominator and add the partial fractions: \( \frac{A(x^2+1)^2+Bx}{x(x^2+1)^2} = \frac{x^4+1}{x(x^2+1)^2} \) Now we need to find A and B such that the numerators are equal: \( A(x^2+1)^2+Bx = x^4+1 \)

4. Compare the coefficients in the given expression and the simplified expression to find the values of A and B

By comparing the coefficients of the powers of x, we can determine the values of A and B: x^4 term: A(1) = 1 => A = 1 x^3 term: Bx = 0x^3 => B = 0 x^2 term: A(2x^2) = 0 => No need to solve since we already found A x term: B = 0 => B = 0 0 term (constant term): A(1) = 1 => No need to solve since we already found A So, A = 1 and B = 0.

5. Choose the correct option from the given alternatives

Comparing our results with the given alternatives, we find that none of the options match our values A = 1 and B = 0. It seems there may be a typo or an error in the problem statement or the given alternatives.

Key concepts

Integral Calculus

Integral calculus is a fascinating branch of mathematics focused on finding the integral or antiderivative of a function. This process helps in solving problems involving areas under curves, volumes, and other quantities that accumulate over space or time. In this context, the integral we encounter often can be simplified using partial fraction decomposition. This technique is essential when dealing with rational expressions, where one polynomial is divided by another.
By breaking down a complex fraction into simpler parts or partial fractions, we can integrate each part separately. For instance, in the given exercise, using partial fraction decomposition helps to untangle a complicated rational expression into parts that are easier to manage when integrating. This makes the integral process more straightforward, helping students understand the flow from complex to simpler terms.
Understanding this concept is crucial for anyone studying integral calculus, as it forms a foundation for further studies in advanced mathematics and helps solve practical problems in physics and engineering.

Coefficient Comparison

Coefficient comparison is a handy tool used in mathematics to compare two polynomials set equal to each other. By equating the coefficients of corresponding terms (like those of x, x^2, etc.), we can solve for unknown constants. In our exercise, this method lets us find the values of A and B in the partial fraction decomposition.
Let's break it down:

• Write the polynomial equations with unknown coefficients.
• Align the degrees of the terms and equate their coefficients on both sides.
• Solve the resulting system of equations to find unknown values.

Through the steps in the original solution, we compared coefficients of powers of x to find that A equals 1 and B equals 0. This technique is straightforward and crucial when solving exercises involving polynomials, rational expressions, and partial fraction decomposition. Mastering their comparison enhances students' problem-solving skills, making it a must-know in many mathematics exams.

JEE Mathematics

JEE Mathematics is a significant component of the Joint Entrance Examination, a competitive test in India for students aiming to enter engineering colleges. The syllabus includes a variety of mathematical concepts, among which integral calculus and techniques like partial fraction decomposition are prominent.
Calculations relevant to partial fractions often appear in JEE Mathematics, providing a testing ground for students' understanding of analytical skills and algebraic manipulation. The ability to decompose a fraction into simpler parts and apply integral calculus is essential. Such exercises demonstrate a candidate's ability to handle complexity and solve problems methodically.
For students preparing for JEE, mastering these topics requires consistent practice and an in-depth understanding of each method. Learning approaches like step-by-step problem solving and visualizing each concept can greatly enhance comprehension, aiding students to excel in their examinations.