How do you find the antiderivative of int 1/(2x^2-x-3) dx?

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Steve M Share
Nov 15, 2016

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

Explanation:

First we complete the square of the denominator:
Let $I = \int \frac{1}{2 {x}^{2} - x - 3} \mathrm{dx}$

$\therefore I = \int \frac{1}{2 \left\{{x}^{2} - \frac{1}{2} x - \frac{3}{2}\right\}} \mathrm{dx}$
$\therefore I = \frac{1}{2} \int \frac{1}{\left\{{\left(x - \frac{1}{4}\right)}^{2} - {\left(\frac{1}{4}\right)}^{2} - 3\right\}} \mathrm{dx}$
$\therefore I = \frac{1}{2} \int \frac{1}{\left\{{\left(x - \frac{1}{4}\right)}^{2} - \frac{1}{16} - 3\right\}} \mathrm{dx}$
$\therefore I = \frac{1}{2} \int \frac{1}{\left\{{\left(x - \frac{1}{4}\right)}^{2} - \frac{49}{16}\right\}} \mathrm{dx}$
$\therefore I = \frac{1}{2} \int \frac{1}{\left\{{\left(x - \frac{1}{4}\right)}^{2} - {\left(\frac{7}{4}\right)}^{2}\right\}} \mathrm{dx}$

Let $u = x - \frac{1}{4} \implies \frac{\mathrm{du}}{\mathrm{dx}} = 1$, Applying this substitution we get:
$\therefore I = \frac{1}{2} \int \frac{1}{{u}^{2} - {\left(\frac{7}{4}\right)}^{2}} \mathrm{du}$

Now a standard integral (that should be learnt if you are studying College Maths) is:
$\int \frac{1}{{x}^{2} - {a}^{2}} \mathrm{dx} = \frac{1}{2 a} \ln | \frac{x - a}{x + a} |$

And so,
$\therefore I = \frac{1}{2} \frac{1}{2 \left(\frac{7}{4}\right)} \ln | \frac{u - \frac{7}{4}}{u + \frac{7}{4}} |$
$\therefore I = \frac{1}{7} \ln | \frac{\left(x - \frac{1}{4}\right) - \frac{7}{4}}{\left(x - \frac{1}{4}\right) + \frac{7}{4}} |$
$\therefore I = \frac{1}{7} \ln | \frac{x - 2}{x + \frac{3}{2}} |$

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