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

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

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

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

- 5 ln (2) + 3 ln (3)

Decimal Notation

- 0.16989 \dots

Solution steps

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

Take the partial fraction of

\frac{x - 1}{x ^{2} + 3 x + 2} : - \frac{2}{x + 1} + \frac{3}{x + 2}

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

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

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

= - 2 ln (2) + 3 (ln (3) - ln (2))

Simplify

- 2 ln (2) + 3 (ln (3) - ln (2)) : - 5 ln (2) + 3 ln (3)

= - 5 ln (2) + 3 ln (3)

\frac{\partial}{\partial x} (ln (x + 7 y + 8 z))

\int e^{2 x} - e^{- x} d x

f (x) = \frac{e ^{x}}{x}, at x = 1

\frac{\partial}{\partial x} (x^{2} y + y^{2} z + x z^{2})

\frac{d}{d x} (\frac{3 x + 2}{5 - 4 x})

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