Question

Find: \( \int \frac{5x}{(x + 1)(x^2 + 9)} dx \).

Answer 1

We use partial fraction decomposition for the integrand \( \frac{5x}{(x + 1)(x^2 + 9)} \). Let $$ \frac{5x}{(x + 1)(x^2 + 9)} = \frac{A}{x + 1} + \frac{Bx + C}{x^2 + 9} $$ Multiplying both sides by \( (x + 1)(x^2 + 9) \), we get: $$ 5x = A(x^2 + 9) + (Bx + C)(x + 1) $$ $$ 5x = Ax^2 + 9A + Bx^2 + Bx + Cx + C $$ $$ 5x = (A + B)x^2 + (B + C)x + (9A + C) $$ Equating the coefficients of \( x^2, x, \) and the constant term, we get the following system of equations: $$ A + B = 0 \quad \cdots (1) $$ $$ B + C = 5 \quad \cdots (2) $$ $$ 9A + C = 0 \quad \cdots (3) $$ From equation (1), \( B = -A \). Substitute \( B = -A \) into equation (2): $$ -A + C = 5 \quad \cdots (4) $$ Now we have two equations with \( A \) and \( C \): $$ 9A + C = 0 \quad \cdots (3) $$ $$ -A + C = 5 \quad \cdots (4) $$ Subtract equation (4) from equation (3): $$ (9A + C) - (-A + C) = 0 - 5 $$ $$ 10A = -5 $$ $$ A = -\frac{1}{2} $$ Now, find \( B \) using \( B = -A \): $$ B = -(-\frac{1}{2}) = \frac{1}{2} $$ Find \( C \) using equation (4): $$ -(-\frac{1}{2}) + C = 5 $$ $$ \frac{1}{2} + C = 5 $$ $$ C = 5 - \frac{1}{2} = \frac{9}{2} $$ So, the partial fraction decomposition is: $$ \frac{5x}{(x + 1)(x^2 + 9)} = \frac{-\frac{1}{2}}{x + 1} + \frac{\frac{1}{2}x + \frac{9}{2}}{x^2 + 9} $$ $$ = -\frac{1}{2(x + 1)} + \frac{x + 9}{2(x^2 + 9)} $$ Now, integrate term by term: $$ \int \frac{5x}{(x + 1)(x^2 + 9)} dx = \int \left( -\frac{1}{2(x + 1)} + \frac{x}{2(x^2 + 9)} + \frac{9}{2(x^2 + 9)} \right) dx $$ $$ = -\frac{1}{2} \int \frac{1}{x + 1} dx + \frac{1}{2} \int \frac{x}{x^2 + 9} dx + \frac{9}{2} \int \frac{1}{x^2 + 9} dx $$ For the second integral, let \( u = x^2 + 9 \), then \( du = 2x dx \), so \( x dx = \frac{1}{2} du \). $$ \int \frac{x}{x^2 + 9} dx = \int \frac{1}{u} \cdot \frac{1}{2} du = \frac{1}{2} \ln |u| = \frac{1}{2} \ln (x^2 + 9) $$ For the third integral, use \( \int \frac{1}{x^2 + a^2} dx = \frac{1}{a} \arctan \left( \frac{x}{a} \right) + C \), with \( a = 3 \). $$ \int \frac{1}{x^2 + 9} dx = \frac{1}{3} \arctan \left( \frac{x}{3} \right) + C $$ Combining the results: $$ \int \frac{5x}{(x + 1)(x^2 + 9)} dx = -\frac{1}{2} \ln |x + 1| + \frac{1}{2} \cdot \frac{1}{2} \ln (x^2 + 9) + \frac{9}{2} \cdot \frac{1}{3} \arctan \left( \frac{x}{3} \right) + C' $$ $$ = -\frac{1}{2} \ln |x + 1| + \frac{1}{4} \ln (x^2 + 9) + \frac{3}{2} \arctan \left( \frac{x}{3} \right) + C $$

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