What is the integral of (y+4)/(y^3+2y^2+2y) ?

The integral of (y+4)/(y^3+2y^2+2y) is -arctan(y+1)+2ln|y|-ln|y^2+2y+2|+C

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integral of (y+4)/(y^3+2y^2+2y)

\int \frac{y + 4}{y ^{3} + 2 y ^{2} + 2 y} d y

- arctan (y + 1) + 2 ln ∣ y ∣ - ln y^{2} + 2 y + 2 + C

Solution steps

\int \frac{y + 4}{y ^{3} + 2 y ^{2} + 2 y} d y

Expand

\frac{y + 4}{y ^{3} + 2 y ^{2} + 2 y} : \frac{y}{y ^{3} + 2 y ^{2} + 2 y} + \frac{4}{y ^{3} + 2 y ^{2} + 2 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{y}{y ^{3} + 2 y ^{2} + 2 y} d y + \int \frac{4}{y ^{3} + 2 y ^{2} + 2 y} d y

\int \frac{y}{y ^{3} + 2 y ^{2} + 2 y} d y = arctan (y + 1)

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

= arctan (y + 1) + 2 ln ∣ y ∣ - ln y^{2} + 2 y + 2 - 2 arctan (y + 1)

Simplify

= - arctan (y + 1) + 2 ln ∣ y ∣ - ln y^{2} + 2 y + 2

Add a constant to the solution

= - arctan (y + 1) + 2 ln ∣ y ∣ - ln y^{2} + 2 y + 2 + C

\frac{d}{d x} (5 x^{3} y)

\frac{t sin ( t )}{1 + t}

\frac{x}{1 + x ^{2}}, \frac{( x ^{2} )}{1 + x ^{3}}

y^{^{'}} = \frac{x y + 2 y}{x + 3}

x (x^{3} + 1) y^{^{'}} + (2 x^{3} - 1) y = \frac{x ^{3} - 2}{x}

What is the integral of (y+4)/(y^3+2y^2+2y) ?
The integral of (y+4)/(y^3+2y^2+2y) is -arctan(y+1)+2ln|y|-ln|y^2+2y+2|+C