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

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

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

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

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

Solution steps

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

Take the partial fraction of

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

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

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

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

= - \frac{1}{y} - arctan (y)

Add a constant to the solution

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

M^{2} (\frac{C}{2} - \frac{M}{3})

y^{^{'}} = 0.002 y - 200, y (0) = 80000

e^{4 x^{2}}

x y^{^{'}} = y + 8 x^{2} sin (x)

x \to 4 - lim (\frac{2}{x - 4})

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