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

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

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integral of 1/(4y^2+1)

\int \frac{1}{4 y ^{2} + 1} d y

\frac{1}{2} arctan (2 y) + C

Solution steps

\int \frac{1}{4 y ^{2} + 1} d y

Apply Integral Substitution

= \int \frac{1}{2 ( u ^{2} + 1 )} 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{1}{u ^{2} + 1} d u

Use the common integral:

\int \frac{1}{u ^{2} + 1} d u = arctan (u)

= \frac{1}{2} arctan (u)

Substitute back

u = 2 y

= \frac{1}{2} arctan (2 y)

Add a constant to the solution

= \frac{1}{2} arctan (2 y) + C

\frac{\partial}{\partial x} (x ln (1 + x + y))

\int_{0}^{1} x d x

e^{x} (7 x^{2} - 14 x + 14)

\frac{4}{x ^{2}} - \frac{1}{x}

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

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