What is the integral of 1/(y^2+9) ?

The integral of 1/(y^2+9) is 1/3 arctan(y/3)+C

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integral of 1/(y^2+9)

\int \frac{1}{y ^{2} + 9} d y

\frac{1}{3} arctan (\frac{y}{3}) + C

Solution steps

\int \frac{1}{y ^{2} + 9} d y

Apply Integral Substitution

= \int \frac{1}{3 ( 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}{3} \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}{3} arctan (u)

Substitute back

u = \frac{y}{3}

= \frac{1}{3} arctan (\frac{y}{3})

Add a constant to the solution

= \frac{1}{3} arctan (\frac{y}{3}) + C

(14 x - x^{2})^{3}

f (x) = \frac{1}{x}, at x = 9

\frac{4}{e ^{2 x}}

x \to 0 lim ((\frac{5}{6})^{\frac{1}{x}})

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

What is the integral of 1/(y^2+9) ?
The integral of 1/(y^2+9) is 1/3 arctan(y/3)+C