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

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

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

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

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

Solution steps

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

Apply u-substitution

: \int \frac{arctan ( u )}{2 ( 1 + u ^{2} )} d u

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

Take the constant out:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

= \frac{1}{2} \cdot \int \frac{arctan ( u )}{1 + u ^{2}} d u

Apply u-substitution

: \int v d v

= \frac{1}{2} \cdot \int v d v

Apply the Power Rule

: \frac{v ^{2}}{2}

= \frac{1}{2} \cdot \frac{v ^{2}}{2}

Substitute back

= \frac{1}{2} \cdot \frac{arctan ^{2} ( 2 x )}{2}

Simplify

\frac{1}{2} \cdot \frac{arctan ^{2} ( 2 x )}{2} : \frac{1}{4} arctan^{2} (2 x)

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

Add a constant to the solution

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

\frac{1}{12 x ^{2}}, x, \frac{x}{10}

16 y^{^{''''}} + 24 y^{^{''}} + 9 y = 0

\frac{\partial}{\partial x} (x^{3} - 3 y^{2})

x \to 4 lim (\frac{1}{x})

n = 0 \sum \infty 6 x^{n}

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