What is the integral of (3x^2sqrt(16+y^2))/y ?

The integral of (3x^2sqrt(16+y^2))/y is (x^3sqrt(16+y^2))/y+C

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integral of (3x^2sqrt(16+y^2))/y

\int \frac{3 x ^{2} 16 + y ^{2}}{y}

\frac{x ^{3} 16 + y ^{2}}{y} + C

Solution steps

\int \frac{3 x ^{2} 16 + y ^{2}}{y} d x

Take the constant out:

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

= \frac{3 16 + y ^{2}}{y} \cdot \int x^{2} d x

Apply the Power Rule

= \frac{3 16 + y ^{2}}{y} \cdot \frac{x ^{3}}{3}

Simplify

\frac{3 16 + y ^{2}}{y} \cdot \frac{x ^{3}}{3} : \frac{x ^{3} 16 + y ^{2}}{y}

= \frac{x ^{3} 16 + y ^{2}}{y}

Add a constant to the solution

= \frac{x ^{3} 16 + y ^{2}}{y} + C

What is the integral of (3x^2sqrt(16+y^2))/y ?
The integral of (3x^2sqrt(16+y^2))/y is (x^3sqrt(16+y^2))/y+C