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

The integral of ((x^2+x-1))/(x^2-x) is ln|x^2-x|+x+C

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integral of ((x^2+x-1))/(x^2-x)

\int \frac{( x ^{2} + x - 1 )}{x ^{2} - x} d x

ln x^{2} - x + x + C

Solution steps

\int \frac{x ^{2} + x - 1}{x ^{2} - x} d x

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

= \int \frac{2 x - 1}{x ^{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 \frac{2 x - 1}{x ^{2} - x} d x + \int 1 d x

\int \frac{2 x - 1}{x ^{2} - x} d x = ln x^{2} - x

\int 1 d x = x

= ln x^{2} - x + x

Add a constant to the solution

= ln x^{2} - x + x + C

f (t) = \frac{cos ( t )}{t}

\frac{\frac{1}{x ^{2}} d u}{d x} - \frac{2 u}{x ^{3}} = - \frac{1}{x ^{2}}

\int (x + \frac{1}{x})^{2} d x

y^{^{'}} = x y^{3}

4 y^{^{''}} + 6 y^{^{'}} - y = 5

What is the integral of ((x^2+x-1))/(x^2-x) ?
The integral of ((x^2+x-1))/(x^2-x) is ln|x^2-x|+x+C