What is the integral of (x^3-3x+1)/(y+sqrt(y)) ?

The integral of (x^3-3x+1)/(y+sqrt(y)) is (1-sqrt(y)}{sqrt(y)(1-y)}(x+\frac{x^4)/4-(3x^2)/2)+C

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integral of (x^3-3x+1)/(y+sqrt(y))

\int \frac{x ^{3} - 3 x + 1}{y + y} d x

\frac{1 - y}{y ( 1 - y )} (x + \frac{x ^{4}}{4} - \frac{3 x ^{2}}{2}) + C

Solution steps

\int \frac{x ^{3} - 3 x + 1}{y + y} d x

Take the constant out:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

= \frac{1}{y + y} \cdot \int 1 + x^{3} - 3 x d x

Simplify

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

= \frac{1 - y}{y ( 1 - y )} \cdot \int 1 + x^{3} - 3 x d x

Apply the Sum Rule:

\int f (x) \pm g (x) d x = \int f (x) d x \pm \int g (x) d x

= \frac{1 - y}{y ( 1 - y )} (\int 1 d x + \int x^{3} d x - \int 3 x d x)

\int 1 d x = x

\int x^{3} d x = \frac{x ^{4}}{4}

\int 3 x d x = \frac{3 x ^{2}}{2}

= \frac{1 - y}{y ( 1 - y )} (x + \frac{x ^{4}}{4} - \frac{3 x ^{2}}{2})

Add a constant to the solution

= \frac{1 - y}{y ( 1 - y )} (x + \frac{x ^{4}}{4} - \frac{3 x ^{2}}{2}) + C

\int (4 e^{- 0.8 x}) d x

\frac{d x}{d t} = e^{- k t}

x^{2}, x^{3}

\frac{\partial}{\partial y} ((x^{2} + y^{2})^{\frac{3}{2}})

\frac{\partial}{\partial x} (arctan (\frac{y}{x + a}))

What is the integral of (x^3-3x+1)/(y+sqrt(y)) ?
The integral of (x^3-3x+1)/(y+sqrt(y)) is (1-sqrt(y)}{sqrt(y)(1-y)}(x+\frac{x^4)/4-(3x^2)/2)+C