What is the integral of (y^2+2y+1)/((y^2+1)^2) ?

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

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integral of (y^2+2y+1)/((y^2+1)^2)

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

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

Solution steps

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

Take the partial fraction of

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

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

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

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

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

Add a constant to the solution

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

4 y^{^{''}} + 28 y^{^{'}} + 49 y = 0

f (t) = e^{t + 9}

x \to + 1 lim (\frac{4}{x - 1})

\frac{d}{d x} (e^{- 4 x} - 4 e^{- 4 x} x)

\frac{d}{d x} (ln (\frac{( x + y )}{( x - y )}))

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