What is the integral of (y^2+38y+361)/((y^2+361)^2) ?

The integral of (y^2+38y+361)/((y^2+361)^2) is 1/19 arctan(y/(19))-(19)/(y^2+361)+C

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integral of (y^2+38y+361)/((y^2+361)^2)

\int \frac{y ^{2} + 38 y + 361}{( y ^{2} + 361 ) ^{2}} d y

\frac{1}{19} arctan (\frac{y}{19}) - \frac{19}{y ^{2} + 361} + C

Solution steps

\int \frac{y ^{2} + 38 y + 361}{( y ^{2} + 361 ) ^{2}} d y

Take the partial fraction of

\frac{y ^{2} + 38 y + 361}{( y ^{2} + 361 ) ^{2}} : \frac{1}{y ^{2} + 361} + \frac{38 y}{( y ^{2} + 361 ) ^{2}}

= \int \frac{1}{y ^{2} + 361} + \frac{38 y}{( y ^{2} + 361 ) ^{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} + 361} d y + \int \frac{38 y}{( y ^{2} + 361 ) ^{2}} d y

\int \frac{1}{y ^{2} + 361} d y = \frac{1}{19} arctan (\frac{y}{19})

\int \frac{38 y}{( y ^{2} + 361 ) ^{2}} d y = - \frac{19}{y ^{2} + 361}

= \frac{1}{19} arctan (\frac{y}{19}) - \frac{19}{y ^{2} + 361}

Add a constant to the solution

= \frac{1}{19} arctan (\frac{y}{19}) - \frac{19}{y ^{2} + 361} + C

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What is the integral of (y^2+38y+361)/((y^2+361)^2) ?
The integral of (y^2+38y+361)/((y^2+361)^2) is 1/19 arctan(y/(19))-(19)/(y^2+361)+C