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

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

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

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

\frac{1}{2} ln ∣ y - 1 ∣ + \frac{1}{2} ln ∣ y + 1 ∣ - \frac{1}{y - 1} + C

Solution steps

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

Expand

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

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

\int \frac{y ^{2}}{( y - 1 ) ^{2} ( y + 1 )} d y = \frac{3}{4} ln ∣ y - 1 ∣ - \frac{1}{2 ( y - 1 )} + \frac{1}{4} ln ∣ y + 1 ∣

\int \frac{1}{( y - 1 ) ^{2} ( y + 1 )} d y = - \frac{1}{4} ln ∣ y - 1 ∣ - \frac{1}{2 ( y - 1 )} + \frac{1}{4} ln ∣ y + 1 ∣

= \frac{3}{4} ln ∣ y - 1 ∣ - \frac{1}{2 ( y - 1 )} + \frac{1}{4} ln ∣ y + 1 ∣ - \frac{1}{4} ln ∣ y - 1 ∣ - \frac{1}{2 ( y - 1 )} + \frac{1}{4} ln ∣ y + 1 ∣

Simplify

\frac{3}{4} ln ∣ y - 1 ∣ - \frac{1}{2 ( y - 1 )} + \frac{1}{4} ln ∣ y + 1 ∣ - \frac{1}{4} ln ∣ y - 1 ∣ - \frac{1}{2 ( y - 1 )} + \frac{1}{4} ln ∣ y + 1 ∣ : \frac{1}{2} ln ∣ y - 1 ∣ + \frac{1}{2} ln ∣ y + 1 ∣ - \frac{1}{y - 1}

= \frac{1}{2} ln ∣ y - 1 ∣ + \frac{1}{2} ln ∣ y + 1 ∣ - \frac{1}{y - 1}

Add a constant to the solution

= \frac{1}{2} ln ∣ y - 1 ∣ + \frac{1}{2} ln ∣ y + 1 ∣ - \frac{1}{y - 1} + C

y^{^{''}} + 2 y^{^{'}} + 4 y = e^{t} cos (t)

\frac{\partial}{\partial z} (y sin (x z))

t \to - 1 lim (\frac{t ^{10} + t ^{5} + 1}{t - 1})

1936 - 16 t^{2}

\frac{x}{x - 5}

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