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

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

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

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

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

Solution steps

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

Take the constant out:

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

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

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

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

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

Add a constant to the solution

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

\int \frac{1}{1 - cos ( x )} d x

\frac{\partial}{\partial u} (u + v^{2})

\int_{0}^{5} (x + 1) d x

\int (θ^{2} - csc^{2} (θ)) d θ

y = x^{2}, y = 7 + x

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