What is the integral of 6y^2+1/(24y^2) ?

The integral of 6y^2+1/(24y^2) is 2y^3-1/(24y)+C

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integral of 6y^2+1/(24y^2)

\int 6 y^{2} + \frac{1}{24 y ^{2}} d y

2 y^{3} - \frac{1}{24 y} + C

Solution steps

\int 6 y^{2} + \frac{1}{24 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 6 y^{2} d y + \int \frac{1}{24 y ^{2}} d y

\int 6 y^{2} d y = 2 y^{3}

\int \frac{1}{24 y ^{2}} d y = - \frac{1}{24 y}

= 2 y^{3} - \frac{1}{24 y}

Add a constant to the solution

= 2 y^{3} - \frac{1}{24 y} + C

What is the integral of 6y^2+1/(24y^2) ?
The integral of 6y^2+1/(24y^2) is 2y^3-1/(24y)+C