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

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

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

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

\frac{1}{2} arctan (2 x) - \frac{1}{8} ln 1 + 4 x^{2} + C

Solution steps

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

Expand

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

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

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

\int \frac{x}{1 + 4 x ^{2}} d x = \frac{1}{8} ln 1 + 4 x^{2}

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

Add a constant to the solution

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

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