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)

\int \frac{7 x ^{3} + 4 x + 4 x ^{2} + 1}{x ^{4} + x ^{2}} d x

\frac{d}{d x} (\frac{- 2 x}{1 - x ^{2}})

\int \frac{4 x}{x - 6} d x

\frac{\partial}{\partial z} (x y + z)

y^{^{'}} = 1 + 4 y^{2}, y (1) = 0

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)