What is the integral of arctan(4t) ?

The integral of arctan(4t) is tarctan(4t)-1/8 ln|16t^2+1|+C

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integral of arctan(4t)

\int arctan (4 t) d t

t arctan (4 t) - \frac{1}{8} ln 16 t^{2} + 1 + C

Solution steps

\int arctan (4 t) d t

Apply Integration By Parts

: t arctan (4 t) - \int \frac{4 t}{16 t ^{2} + 1} d t

= t arctan (4 t) - \int \frac{4 t}{16 t ^{2} + 1} d t

\int \frac{4 t}{16 t ^{2} + 1} d t = \frac{1}{8} ln 16 t^{2} + 1

= t arctan (4 t) - \frac{1}{8} ln 16 t^{2} + 1

Add a constant to the solution

= t arctan (4 t) - \frac{1}{8} ln 16 t^{2} + 1 + C

\frac{\partial}{\partial y} (- y)

\frac{ln ( x )}{x ^{6}}

x \to \infty lim (0.99)

x y^{^{'}} = e^{- y}

\int e^{- 8 x} d x

What is the integral of arctan(4t) ?
The integral of arctan(4t) is tarctan(4t)-1/8 ln|16t^2+1|+C