What is the integral of arctan(2t) ?

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

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integral of arctan(2t)

\int arctan (2 t) d t

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

Solution steps

\int arctan (2 t) d t

Apply Integration By Parts

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

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

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

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

Add a constant to the solution

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

x \to + 4 lim (3)

y^{^{''}} - 3 y^{^{'}} + 4 y = 0

\frac{d}{d x} (\frac{( 4 x - 2 ) ^{4}}{3 x ^{2} + 7})

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

x y^{^{'}} = y + (x y)^{\frac{1}{2}}

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