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

The integral of (2x+1)/(x^2+x-1) is ln|4x^2+4x-4|+C

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

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

ln 4 x^{2} + 4 x - 4 + C

Solution steps

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

Expand

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

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

\int \frac{2 x}{x ^{2} + x - 1} d x = ln 4 x^{2} + 4 x - 4 + \frac{1}{5} (ln \frac{2 x + 1}{5} + 1 - ln \frac{2 x + 1}{5} - 1)

\int \frac{1}{x ^{2} + x - 1} d x = - \frac{1}{5} (ln \frac{2 x + 1}{5} + 1 - ln \frac{2 x + 1}{5} - 1)

= ln 4 x^{2} + 4 x - 4 + \frac{1}{5} (ln \frac{2 x + 1}{5} + 1 - ln \frac{2 x + 1}{5} - 1) - \frac{1}{5} (ln \frac{2 x + 1}{5} + 1 - ln \frac{2 x + 1}{5} - 1)

Add similar elements:

\frac{1}{5} (ln \frac{2 x + 1}{5} + 1 - ln \frac{2 x + 1}{5} - 1) - \frac{1}{5} (ln \frac{2 x + 1}{5} + 1 - ln \frac{2 x + 1}{5} - 1) = 0

= ln 4 x^{2} + 4 x - 4

Add a constant to the solution

= ln 4 x^{2} + 4 x - 4 + C

x \to 3 + lim (\frac{2 x}{x - 3})

\frac{\partial}{\partial x} (e^{- 3 x} sin (2 t - 3 x))

\frac{1}{4 + 3 x}

\int \frac{x ^{2}}{x ^{3} - 5} d x

\frac{1}{2 x}, at 2

What is the integral of (2x+1)/(x^2+x-1) ?
The integral of (2x+1)/(x^2+x-1) is ln|4x^2+4x-4|+C