Problems on Techniques of Integration

We have

$$\frac{3x-1}{x^2 + x -2} = \frac{3x-1}{(x-1)(x+2)} = \frac{A}{x-1} + \frac{B}{x + 2}$$

which implies $A = \displaystyle \frac{2}{3}$ and $B = \displaystyle \frac{7}{3}$, or

$$\frac{3x-1}{x^2 + x -2} = \frac{1}{3} \left[\frac{2}{x-1} + \frac{7}{x + 2}\right]\;.$$

Hence

$$\int \frac{3x-1}{x^2 + x -2}dx = \frac{2}{3} \int \frac{1}{x-1}dx + \frac{7}{3}\int \frac{1}{x + 2}dx$$

which implies

$$\int \frac{x+1}{x^2 + 4}dx = \frac{2}{3} \ln\vert x-1\vert + \frac{7}{3} \ln\vert x+2\vert + C\;.$$

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