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

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

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integral of 1/(y^2+4)

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

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

Solution steps

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

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

Add a constant to the solution

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

z

x (x^{2} + 6) (x + 9)

f (x) = e^{4 x} cos (π x), at x = 0

cos (3 x)

\int x^{9} + 8 d x

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