What is the integral of (y^2+36y+324)/((y^2+324)^2) ?

The integral of (y^2+36y+324)/((y^2+324)^2) is 1/18 arctan(y/(18))-(18)/(y^2+324)+C

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integral of (y^2+36y+324)/((y^2+324)^2)

\int \frac{y ^{2} + 36 y + 324}{( y ^{2} + 324 ) ^{2}} d y

\frac{1}{18} arctan (\frac{y}{18}) - \frac{18}{y ^{2} + 324} + C

Solution steps

\int \frac{y ^{2} + 36 y + 324}{( y ^{2} + 324 ) ^{2}} d y

Take the partial fraction of

\frac{y ^{2} + 36 y + 324}{( y ^{2} + 324 ) ^{2}} : \frac{1}{y ^{2} + 324} + \frac{36 y}{( y ^{2} + 324 ) ^{2}}

= \int \frac{1}{y ^{2} + 324} + \frac{36 y}{( y ^{2} + 324 ) ^{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} + 324} d y + \int \frac{36 y}{( y ^{2} + 324 ) ^{2}} d y

\int \frac{1}{y ^{2} + 324} d y = \frac{1}{18} arctan (\frac{y}{18})

\int \frac{36 y}{( y ^{2} + 324 ) ^{2}} d y = - \frac{18}{y ^{2} + 324}

= \frac{1}{18} arctan (\frac{y}{18}) - \frac{18}{y ^{2} + 324}

Add a constant to the solution

= \frac{1}{18} arctan (\frac{y}{18}) - \frac{18}{y ^{2} + 324} + C

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What is the integral of (y^2+36y+324)/((y^2+324)^2) ?
The integral of (y^2+36y+324)/((y^2+324)^2) is 1/18 arctan(y/(18))-(18)/(y^2+324)+C