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

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

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

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

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

Solution steps

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

Expand

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

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

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

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

Add a constant to the solution

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

\frac{d}{d x} (cos (x) + cos (y))

\frac{d}{d x} (\frac{x}{3 + x})

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

\frac{d y}{d t} + y = 2

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