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

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

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integral of (1/(sqrt(y-2))+3y^2)

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

2 y - 2 + y^{3} + C

Solution steps

\int \frac{1}{y - 2} + 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 \frac{1}{y - 2} d y + \int 3 y^{2} d y

\int \frac{1}{y - 2} d y = 2 y - 2

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

= 2 y - 2 + y^{3}

Add a constant to the solution

= 2 y - 2 + y^{3} + C

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