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

The integral from 1 to 2 of (x-3)/(x^3+x^2) is 8ln(2)-3/2-4ln(3)

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

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

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

Decimal

- 0.34927 \dots

Solution steps

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

Take the partial fraction of

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

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

\int_{1}^{2} \frac{4}{x} d x = 4 ln (2)

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

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

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

Simplify

4 ln (2) - \frac{3}{2} - 4 (ln (3) - ln (2)) : 8 ln (2) - \frac{3}{2} - 4 ln (3)

= 8 ln (2) - \frac{3}{2} - 4 ln (3)

What is the integral from 1 to 2 of (x-3)/(x^3+x^2) ?
The integral from 1 to 2 of (x-3)/(x^3+x^2) is 8ln(2)-3/2-4ln(3)