What is the integral of (3x^3-sqrt(2x-1)) ?

The integral of (3x^3-sqrt(2x-1)) is (3x^4)/4-1/3 (2x-1)^{3/2}+C

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integral of (3x^3-sqrt(2x-1))

\int (3 x^{3} - 2 x - 1) d x

\frac{3 x ^{4}}{4} - \frac{1}{3} (2 x - 1)^{\frac{3}{2}} + C

Solution steps

\int 3 x^{3} - 2 x - 1 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 3 x^{3} d x - \int 2 x - 1 d x

\int 3 x^{3} d x = \frac{3 x ^{4}}{4}

\int 2 x - 1 d x = \frac{1}{3} (2 x - 1)^{\frac{3}{2}}

= \frac{3 x ^{4}}{4} - \frac{1}{3} (2 x - 1)^{\frac{3}{2}}

Add a constant to the solution

= \frac{3 x ^{4}}{4} - \frac{1}{3} (2 x - 1)^{\frac{3}{2}} + C

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What is the integral of (3x^3-sqrt(2x-1)) ?
The integral of (3x^3-sqrt(2x-1)) is (3x^4)/4-1/3 (2x-1)^{3/2}+C