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

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

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

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

\frac{1}{2} ln x + \frac{1}{2} + \frac{1}{2 x + 1} + C

Solution steps

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

Expand

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

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

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

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

Simplify

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

= \frac{1}{2} ln x + \frac{1}{2} + \frac{1}{2 x + 1}

Add a constant to the solution

= \frac{1}{2} ln x + \frac{1}{2} + \frac{1}{2 x + 1} + C

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