What is the integral of (y^2+5)/(y^2+9) ?

The integral of (y^2+5)/(y^2+9) is -4/3 arctan(y/3)+y+C

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integral of (y^2+5)/(y^2+9)

\int \frac{y ^{2} + 5}{y ^{2} + 9} d y

- \frac{4}{3} arctan (\frac{y}{3}) + y + C

Solution steps

\int \frac{y ^{2} + 5}{y ^{2} + 9} d y

\frac{y ^{2} + 5}{y ^{2} + 9} = - \frac{4}{y ^{2} + 9} + 1

= \int - \frac{4}{y ^{2} + 9} + 1 d y

Apply the Sum Rule:

\int f (x) \pm g (x) d x = \int f (x) d x \pm \int g (x) d x

= - \int \frac{4}{y ^{2} + 9} d y + \int 1 d y

\int \frac{4}{y ^{2} + 9} d y = \frac{4}{3} arctan (\frac{y}{3})

\int 1 d y = y

= - \frac{4}{3} arctan (\frac{y}{3}) + y

Add a constant to the solution

= - \frac{4}{3} arctan (\frac{y}{3}) + y + C

\int \frac{3 x sin ^{2} ( x )}{cos ^{4} ( x )} d x

\int \frac{x ^{2} + 1}{x ^{3} - x} d x

\int \frac{- x ^{2}}{4} d x

\int 4 - 4 x d x

What is the integral of (y^2+5)/(y^2+9) ?
The integral of (y^2+5)/(y^2+9) is -4/3 arctan(y/3)+y+C