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

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

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

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

ln (3) - ln (2)

Solution steps

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

Take the partial fraction of

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

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

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

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

= ln (3) - ln (2)

\int_{0}^{1} 6 x^{2} - 6 x^{3} d x

\int_{0}^{4} x^{2} e^{- x} d x

\int_{1}^{2} x^{2} + 1 d x

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

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