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

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

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integral from 2 to 4 of (x+1)/(x^2-x)

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

- ln (2) + 2 ln (3)

Decimal Notation

1.50407 \dots

Solution steps

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

Take the partial fraction of

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

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

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

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

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

\int_{0}^{x} \frac{x ^{4}}{1 - x ^{4}} d x

\int_{0}^{e} \frac{x}{x ^{2} + e} d x

\int_{4}^{6} \frac{6}{x ^{4}} d x

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

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

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