What is the integral of (-4x+4)/(5(4x^2-4x+5)) ?

The integral of (-4x+4)/(5(4x^2-4x+5)) is 1/10 (-ln|4x^2-4x+5|+arctan(x-1/2))+C

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integral of (-4x+4)/(5(4x^2-4x+5))

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

\frac{1}{10} (- ln 4 x^{2} - 4 x + 5 + arctan (x - \frac{1}{2})) + C

Solution steps

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

Take the constant out:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

= \frac{1}{5} \cdot \int \frac{- 4 x + 4}{4 x ^{2} - 4 x + 5} d x

Expand

\frac{- 4 x + 4}{4 x ^{2} - 4 x + 5} : - \frac{4 x}{4 x ^{2} - 4 x + 5} + \frac{4}{4 x ^{2} - 4 x + 5}

Apply the Sum Rule:

\int f (x) \pm g (x) d x = \int f (x) d x \pm \int g (x) d x

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

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

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

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

Simplify

\frac{1}{5} (- 2 (\frac{1}{4} ln 4 x^{2} - 4 x + 5 + \frac{1}{4} arctan (x - \frac{1}{2})) + arctan (x - \frac{1}{2})) : \frac{1}{10} (- ln 4 x^{2} - 4 x + 5 + arctan (x - \frac{1}{2}))

= \frac{1}{10} (- ln 4 x^{2} - 4 x + 5 + arctan (x - \frac{1}{2}))

Add a constant to the solution

= \frac{1}{10} (- ln 4 x^{2} - 4 x + 5 + arctan (x - \frac{1}{2})) + C

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What is the integral of (-4x+4)/(5(4x^2-4x+5)) ?
The integral of (-4x+4)/(5(4x^2-4x+5)) is 1/10 (-ln|4x^2-4x+5|+arctan(x-1/2))+C