What is the integral of 2arctan(3y) ?

The integral of 2arctan(3y) is 2(yarctan(3y)-1/6 ln|9y^2+1|)+C

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integral of 2arctan(3y)

\int 2 arctan (3 y) d y

2 (y arctan (3 y) - \frac{1}{6} ln 9 y^{2} + 1) + C

Solution steps

\int 2 arctan (3 y) d y

Take the constant out:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

= 2 \cdot \int arctan (3 y) d y

Apply Integration By Parts

: y arctan (3 y) - \int \frac{3 y}{9 y ^{2} + 1} d y

= 2 (y arctan (3 y) - \int \frac{3 y}{9 y ^{2} + 1} d y)

\int \frac{3 y}{9 y ^{2} + 1} d y = \frac{1}{6} ln 9 y^{2} + 1

= 2 (y arctan (3 y) - \frac{1}{6} ln 9 y^{2} + 1)

Add a constant to the solution

= 2 (y arctan (3 y) - \frac{1}{6} ln 9 y^{2} + 1) + C

\frac{\partial}{\partial x} (9 + x ln (x y - 5))

\int_{0}^{\infty} sin (9 x) d x

\frac{d}{d x} (sin (x) - x cos (x))

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

\frac{d}{d x} (\frac{2}{2 x + 3})

What is the integral of 2arctan(3y) ?
The integral of 2arctan(3y) is 2(yarctan(3y)-1/6 ln|9y^2+1|)+C