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

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

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

integral

\frac{1}{( 1 + 4 y ^{2} )}

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

Solution steps

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

Apply Integral Substitution

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

Use the common integral:

\int \frac{1}{1 + u ^{2}} 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{- 16 - 6 x ^{4}}{x ^{2}}

\frac{1}{( 1 + 4 \frac{x}{l} )}

\frac{2 x}{x ^{2}}

\frac{x ^{4} + x ^{2} + 1}{( x + 1 )}

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

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