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

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

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

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

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

Solution steps

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

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 t )}{2}

Simplify

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

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

Add a constant to the solution

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

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