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

The integral of ((1-x))/((x^2-x+4)) is -1/2 ln|4x^2-4x+16|+1/(sqrt(15))arctan(1/(sqrt(15))(2x-1))+C

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

integral

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

- \frac{1}{2} ln 4 x^{2} - 4 x + 16 + \frac{1}{15} arctan (\frac{1}{15} (2 x - 1)) + C

Solution steps

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

Expand

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

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

\int \frac{1}{x ^{2} - x + 4} d x = \frac{2}{15} arctan (\frac{2 x - 1}{15})

\int \frac{x}{x ^{2} - x + 4} d x = \frac{1}{2} ln 4 x^{2} - 4 x + 16 + \frac{1}{15} arctan (\frac{2 x - 1}{15})

= \frac{2}{15} arctan (\frac{2 x - 1}{15}) - (\frac{1}{2} ln 4 x^{2} - 4 x + 16 + \frac{1}{15} arctan (\frac{2 x - 1}{15}))

Simplify

\frac{2}{15} arctan (\frac{2 x - 1}{15}) - (\frac{1}{2} ln 4 x^{2} - 4 x + 16 + \frac{1}{15} arctan (\frac{2 x - 1}{15})) : - \frac{1}{2} ln 4 x^{2} - 4 x + 16 + \frac{1}{15} arctan (\frac{1}{15} (2 x - 1))

= - \frac{1}{2} ln 4 x^{2} - 4 x + 16 + \frac{1}{15} arctan (\frac{1}{15} (2 x - 1))

Add a constant to the solution

= - \frac{1}{2} ln 4 x^{2} - 4 x + 16 + \frac{1}{15} arctan (\frac{1}{15} (2 x - 1)) + C

\frac{1}{v ( v + 1 )}

e^{x} e^{2 x} + 6 e^{x} + 10

1 - cos (2 x)

2 e^{2} x

lo g_{10} (\frac{2}{y})

What is the integral of ((1-x))/((x^2-x+4)) ?
The integral of ((1-x))/((x^2-x+4)) is -1/2 ln|4x^2-4x+16|+1/(sqrt(15))arctan(1/(sqrt(15))(2x-1))+C