Question

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

Hint:

• When dealing with a quadratic factor \( ax^2+b \), the numerator in partial fractions should be \( Bx+C \).
• To integrate \( \frac{x}{x^2+a^2} \), always create the derivative of the denominator in the numerator by multiplying/dividing by 2.

Answer 1

Concept:

• Method of Partial Fractions for rational functions.
• Standard integrals: \( \int \frac{1}{x} dx = \log|x| \), \( \int \frac{f'(x)}{f(x)} dx = \log|f(x)| \), and \( \int \frac{1}{x^2+a^2} dx = \frac{1}{a} \tan^{-1}\frac{x}{a} \).

Step 1: Decompose the integrand into partial fractions
Let \( \frac{x}{(x-1)(x^2+4)} = \frac{A}{x-1} + \frac{Bx + C}{x^2+4} \)
Multiplying by the denominator: \( x = A(x^2+4) + (Bx+C)(x-1) \)
Put \( x = 1 \): \( 1 = A(1+4) \implies A = \frac{1}{5} \)
Comparing coefficients of \( x^2 \): \( 0 = A + B \implies B = -A = -\frac{1}{5} \)
Comparing constant terms: \( 0 = 4A - C \implies C = 4A = \frac{4}{5} \)

Step 2: Set up the integral with substituted partial fractions
\( \int \frac{x}{(x-1)(x^2+4)} dx = \int \left[ \frac{1}{5(x-1)} + \frac{-\frac{1}{5}x + \frac{4}{5}}{x^2+4} \right] dx \)
\( = \frac{1}{5} \int \frac{1}{x-1} dx - \frac{1}{5} \int \frac{x}{x^2+4} dx + \frac{4}{5} \int \frac{1}{x^2+4} dx \)

Step 3: Integrate each term
\( I_1 = \frac{1}{5} \log|x-1| \)
\( I_2 = -\frac{1}{5} \times \frac{1}{2} \int \frac{2x}{x^2+4} dx = -\frac{1}{10} \log(x^2+4) \)
\( I_3 = \frac{4}{5} \times \frac{1}{2} \tan^{-1}\frac{x}{2} = \frac{2}{5} \tan^{-1}\frac{x}{2} \)
Final answer: \( \frac{1}{5} \log|x-1| - \frac{1}{10} \log(x^2+4) + \frac{2}{5} \tan^{-1}\frac{x}{2} + C \)