What is the integral from 1 to 3 of (x-6)/(x^2(x+3)) ?

The integral from 1 to 3 of (x-6)/(x^2(x+3)) is ln(3)-4/3-ln(3/2)

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integral from 1 to 3 of (x-6)/(x^2(x+3))

\int_{1}^{3} \frac{x - 6}{x ^{2} ( x + 3 )} d x

ln (3) - \frac{4}{3} - ln (\frac{3}{2})

Decimal Notation

- 0.64018 \dots

Solution steps

\int_{1}^{3} \frac{x - 6}{x ^{2} ( x + 3 )} d x

Take the partial fraction of

\frac{x - 6}{x ^{2} ( x + 3 )} : \frac{1}{x} - \frac{2}{x ^{2}} - \frac{1}{x + 3}

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

\int_{1}^{3} \frac{1}{x} d x = ln (3)

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

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

= ln (3) - \frac{4}{3} - ln (\frac{3}{2})

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What is the integral from 1 to 3 of (x-6)/(x^2(x+3)) ?
The integral from 1 to 3 of (x-6)/(x^2(x+3)) is ln(3)-4/3-ln(3/2)