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

y = 0, y = 2 x^{2}, y = 0, [0, 1]

n = 3 \sum \infty \frac{1}{ln ( n )}

\int \frac{1}{( x - 5 ) ( x - 2 )} d x

\frac{\partial}{\partial x} ((e^{x} + e^{y})^{5})

\int_{1}^{2} (x - 1) d x

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