Q1817: Find: $\int\frac{x}{(x-1)(x^{2}+4)}dx$...

Class CBSE Class 12 Mathematics Integrals Q #1817

COMPETENCY BASED

APPLY

5 Marks 2026 AISSCE(Board Exam) LA

Find: $\int\frac{x}{(x-1)(x^{2}+4)}dx$

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Detailed Solution

Step 1: Partial Fraction Decomposition

We express the integrand as a sum of partial fractions: $$ \frac{x}{(x-1)(x^2+4)} = \frac{A}{x-1} + \frac{Bx+C}{x^2+4} $$ Multiplying both sides by $(x-1)(x^2+4)$, we get: $$ x = A(x^2+4) + (Bx+C)(x-1) $$

Step 2: Solving for Constants

Setting $x=1$: $$ 1 = A(1+4) \implies 5A = 1 \implies A = \frac{1}{5} $$ Comparing coefficients of $x^2$: $0 = A + B \implies B = -A = -\frac{1}{5}$. Comparing constants: $0 = 4A - C \implies C = 4A = \frac{4}{5}$.

Step 3: Integration

The integral becomes: $$ \int \left( \frac{1/5}{x-1} + \frac{-1/5x + 4/5}{x^2+4} \right) dx $$ $$ = \frac{1}{5} \ln|x-1| - \frac{1}{10} \int \frac{2x}{x^2+4} dx + \frac{4}{5} \int \frac{1}{x^2+2^2} dx $$ $$ = \frac{1}{5} \ln|x-1| - \frac{1}{10} \ln(x^2+4) + \frac{4}{5} \cdot \frac{1}{2} \tan^{-1}\left(\frac{x}{2}\right) + C $$

Final Answer: \frac{1}{5} \ln|x-1| - \frac{1}{10} \ln(x^2+4) + \frac{2}{5} \tan^{-1}\left(\frac{x}{2}\right) + C

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Pedagogical Audit

Bloom's Analysis: This is an APPLY question because the student must select and execute the appropriate integration technique (Partial Fractions) to solve a non-standard rational function.

Knowledge Dimension: PROCEDURAL

Justification: The problem requires a multi-step algorithmic process involving algebraic decomposition followed by standard integration formulas.

Syllabus Audit: In the context of CBSE Class 12, this is classified as COMPETENCY. It tests the student's ability to handle integration of rational functions where the denominator contains irreducible quadratic factors, a core competency in the Calculus unit.

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