What is the integral of (2y-3y^4-y^3) 1/y ?

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

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integral of (2y-3y^4-y^3) 1/y

\int (2 y - 3 y^{4} - y^{3}) \frac{1}{y} d y

2 y - \frac{3 y ^{4}}{4} - \frac{y ^{3}}{3} + C

Solution steps

\int (2 y - 3 y^{4} - y^{3}) \frac{1}{y} d y

Simplify

(2 y - 3 y^{4} - y^{3}) \frac{1}{y} : 2 - 3 y^{3} - y^{2}

= \int 2 - 3 y^{3} - y^{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 2 d y - \int 3 y^{3} d y - \int y^{2} d y

\int 2 d y = 2 y

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

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

= 2 y - \frac{3 y ^{4}}{4} - \frac{y ^{3}}{3}

Add a constant to the solution

= 2 y - \frac{3 y ^{4}}{4} - \frac{y ^{3}}{3} + C

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(1 + x^{2}) d y - (1 + 4 y^{2}) d x = 0

\int (x^{2} + 6) \cdot e^{(- x)} d x

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