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

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

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

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

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

Solution steps

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

Expand

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

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

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

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

Add a constant to the solution

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

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