What is the integral of (y^2+20y+100)/((y^2+100)^2) ?

The integral of (y^2+20y+100)/((y^2+100)^2) is 1/10 arctan(y/(10))-(10)/(y^2+100)+C

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integral of (y^2+20y+100)/((y^2+100)^2)

\int \frac{y ^{2} + 20 y + 100}{( y ^{2} + 100 ) ^{2}} d y

\frac{1}{10} arctan (\frac{y}{10}) - \frac{10}{y ^{2} + 100} + C

Solution steps

\int \frac{y ^{2} + 20 y + 100}{( y ^{2} + 100 ) ^{2}} d y

Take the partial fraction of

\frac{y ^{2} + 20 y + 100}{( y ^{2} + 100 ) ^{2}} : \frac{1}{y ^{2} + 100} + \frac{20 y}{( y ^{2} + 100 ) ^{2}}

= \int \frac{1}{y ^{2} + 100} + \frac{20 y}{( y ^{2} + 100 ) ^{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} + 100} d y + \int \frac{20 y}{( y ^{2} + 100 ) ^{2}} d y

\int \frac{1}{y ^{2} + 100} d y = \frac{1}{10} arctan (\frac{y}{10})

\int \frac{20 y}{( y ^{2} + 100 ) ^{2}} d y = - \frac{10}{y ^{2} + 100}

= \frac{1}{10} arctan (\frac{y}{10}) - \frac{10}{y ^{2} + 100}

Add a constant to the solution

= \frac{1}{10} arctan (\frac{y}{10}) - \frac{10}{y ^{2} + 100} + C

y^{^{''}} - 8 y^{^{'}} + 19 y = 0

\frac{\partial}{\partial x} (\frac{1}{x ^{2} + a ^{2}})

y = (x^{3} - 25 x)^{12}, (- 5, 0)

\int a + 18

\frac{d}{d x} (sin (7 π x))

What is the integral of (y^2+20y+100)/((y^2+100)^2) ?
The integral of (y^2+20y+100)/((y^2+100)^2) is 1/10 arctan(y/(10))-(10)/(y^2+100)+C