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

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

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

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

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

Solution steps

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

Apply Integral Substitution

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

Simplify

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

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

Add a constant to the solution

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

\frac{d}{d x} (x^{3} - x + 1)

\int (cosh (x))^{3} d x

\frac{\partial}{\partial x} (x^{\frac{1}{2}} + \frac{Y ( x )}{2})

f (x) = \frac{x ^{2} - 1}{x ^{2} + x + 1}

k = 2 \sum \infty \frac{2}{k ^{2} - 1}

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