What is the integral of (y^2+6y+9)/((y^2+9)^2) ?

The integral of (y^2+6y+9)/((y^2+9)^2) is 1/3 arctan(y/3)-3/(y^2+9)+C

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integral of (y^2+6y+9)/((y^2+9)^2)

\int \frac{y ^{2} + 6 y + 9}{( y ^{2} + 9 ) ^{2}} d y

\frac{1}{3} arctan (\frac{y}{3}) - \frac{3}{y ^{2} + 9} + C

Solution steps

\int \frac{y ^{2} + 6 y + 9}{( y ^{2} + 9 ) ^{2}} d y

Take the partial fraction of

\frac{y ^{2} + 6 y + 9}{( y ^{2} + 9 ) ^{2}} : \frac{1}{y ^{2} + 9} + \frac{6 y}{( y ^{2} + 9 ) ^{2}}

= \int \frac{1}{y ^{2} + 9} + \frac{6 y}{( y ^{2} + 9 ) ^{2}} 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{1}{y ^{2} + 9} d y + \int \frac{6 y}{( y ^{2} + 9 ) ^{2}} d y

\int \frac{1}{y ^{2} + 9} d y = \frac{1}{3} arctan (\frac{y}{3})

\int \frac{6 y}{( y ^{2} + 9 ) ^{2}} d y = - \frac{3}{y ^{2} + 9}

= \frac{1}{3} arctan (\frac{y}{3}) - \frac{3}{y ^{2} + 9}

Add a constant to the solution

= \frac{1}{3} arctan (\frac{y}{3}) - \frac{3}{y ^{2} + 9} + C

f (x) = 3 x - 1, at x = 5

n = 1 \sum \infty \frac{n !}{( 2 n ) !}

x^{4} + \frac{4}{3} x^{3} - 4 x^{2} - \frac{4}{3}

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

f (x) = 5 x^{3} - 5 x, (- 1, 0)

What is the integral of (y^2+6y+9)/((y^2+9)^2) ?
The integral of (y^2+6y+9)/((y^2+9)^2) is 1/3 arctan(y/3)-3/(y^2+9)+C