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

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

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integral of (x^3+1)/y

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

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

Solution steps

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

Take the constant out:

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

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

\int 1 d x = x

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

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

Add a constant to the solution

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

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