What is the integral from 0 to 3 of (x-1)/(x^2-3x-4) ?

The integral from 0 to 3 of (x-1)/(x^2-3x-4) is -2/5 ln(2)

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integral from 0 to 3 of (x-1)/(x^2-3x-4)

\int_{0}^{3} \frac{x - 1}{x ^{2} - 3 x - 4} d x

- \frac{2}{5} ln (2)

Decimal

- 0.27725 \dots

Solution steps

\int_{0}^{3} \frac{x - 1}{x ^{2} - 3 x - 4} d x

Take the partial fraction of

\frac{x - 1}{x ^{2} - 3 x - 4} : \frac{2}{5 ( x + 1 )} + \frac{3}{5 ( x - 4 )}

= \int_{0}^{3} \frac{2}{5 ( x + 1 )} + \frac{3}{5 ( x - 4 )} d x

Apply the Sum Rule:

\int f (x) \pm g (x) d x = \int f (x) d x \pm \int g (x) d x

= \int_{0}^{3} \frac{2}{5 ( x + 1 )} d x + \int_{0}^{3} \frac{3}{5 ( x - 4 )} d x

\int_{0}^{3} \frac{2}{5 ( x + 1 )} d x = \frac{4}{5} ln (2)

\int_{0}^{3} \frac{3}{5 ( x - 4 )} d x = - \frac{6}{5} ln (2)

= \frac{4}{5} ln (2) - \frac{6}{5} ln (2)

Simplify

\frac{4}{5} ln (2) - \frac{6}{5} ln (2) : - \frac{2}{5} ln (2)

= - \frac{2}{5} ln (2)

What is the integral from 0 to 3 of (x-1)/(x^2-3x-4) ?
The integral from 0 to 3 of (x-1)/(x^2-3x-4) is -2/5 ln(2)