What is the integral of (6^2y-3xsqrt(y)) ?

The integral of (6^2y-3xsqrt(y)) is 36yx-(3sqrt(y)x^2)/2+C

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integral of (6^2y-3xsqrt(y))

\int (6^{2} y - 3 x y) d x

36 y x - \frac{3 y x ^{2}}{2} + C

Solution steps

\int 6^{2} y - 3 x y 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

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

\int 6^{2} y d x = 36 y x

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

= 36 y x - \frac{3 y x ^{2}}{2}

Add a constant to the solution

= 36 y x - \frac{3 y x ^{2}}{2} + C

What is the integral of (6^2y-3xsqrt(y)) ?
The integral of (6^2y-3xsqrt(y)) is 36yx-(3sqrt(y)x^2)/2+C