Problems on Techniques of Integration

We have

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

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

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

Hence

\begin{displaymath}\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\end{displaymath}

which implies

\begin{displaymath}\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\;.\end{displaymath}

Detailed Answer.


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