What is the integral of (-4x+1)/(4x^2-2x+5) ?

The integral of (-4x+1)/(4x^2-2x+5) is -1/2 ln|16x^2-8x+20|+C

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integral of (-4x+1)/(4x^2-2x+5)

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

- \frac{1}{2} ln 16 x^{2} - 8 x + 20 + C

Solution steps

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

Expand

\frac{- 4 x + 1}{4 x ^{2} - 2 x + 5} : - \frac{4 x}{4 x ^{2} - 2 x + 5} + \frac{1}{4 x ^{2} - 2 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

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

\int \frac{4 x}{4 x ^{2} - 2 x + 5} d x = \frac{1}{2} ln 16 x^{2} - 8 x + 20 + \frac{1}{19} arctan (\frac{4 x - 1}{19})

\int \frac{1}{4 x ^{2} - 2 x + 5} d x = \frac{1}{19} arctan (\frac{4 x - 1}{19})

= - (\frac{1}{2} ln 16 x^{2} - 8 x + 20 + \frac{1}{19} arctan (\frac{4 x - 1}{19})) + \frac{1}{19} arctan (\frac{4 x - 1}{19})

Simplify

- (\frac{1}{2} ln 16 x^{2} - 8 x + 20 + \frac{1}{19} arctan (\frac{4 x - 1}{19})) + \frac{1}{19} arctan (\frac{4 x - 1}{19}) : - \frac{1}{2} ln 16 x^{2} - 8 x + 20

= - \frac{1}{2} ln 16 x^{2} - 8 x + 20

Add a constant to the solution

= - \frac{1}{2} ln 16 x^{2} - 8 x + 20 + C

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What is the integral of (-4x+1)/(4x^2-2x+5) ?
The integral of (-4x+1)/(4x^2-2x+5) is -1/2 ln|16x^2-8x+20|+C