Question

Use partial fractions to find the indefinite integral. $$ \int \frac{3}{x^{2}+x-2} d x $$

Short answer

The solution to \(\int \frac{3}{x^{2}+x-2} dx\) is \(ln |x - 1| - ln |x+2| + C\).

Step-by-step solution

Factor the quadratic

Firstly, it's important to factor the denominator in the integral. To do this, find two numbers that multiplied are equal to -2 (c in the quadratic equation), and added are equal to 1 (b in the quadratic equation). The equation \(x^2+x-2\) factors into \((x-1)(x+2)\). So, we rewrite the integral: \[\int \frac{3}{(x-1)(x+2)} dx \]

Decomposition into Partial Fractions

Now, we need to decompose the fraction into partial fractions. We write \[\frac{3}{(x-1)(x+2)} = \frac{A}{x-1} + \frac{B}{x+2}\]Cross-multiplying gives:\[3 = A(x + 2) + B(x - 1)\]Setting \(x = 1\), we get \(A=1\). Setting \(x = -2\), we find \(B=1\). Therefore, we may write the integral: \[\int \frac{1}{x-1} dx + \int \frac{1}{x+2} dx \]

Integral computation

The integrals we have are now straightforward to compute. The antiderivative of \(\frac{1}{x-a}\) is \(ln |x-a|\). So the integral of \(\frac{1}{x-1}\) is \(ln |x - 1|\) and the integral of \(\frac{1}{x+2}\) is \(ln |x+2|\). Thus the integral of interest is: \[\int \frac{1}{x-1} dx + \int \frac{1}{x+2} dx = ln |x - 1| - ln |x+2| + C \]where \(C\) is the constant of integration.