What is the integral of (3x^3+4x-7) ?

The integral of (3x^3+4x-7) is (3x^4)/4+2x^2-7x+C

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integral of (3x^3+4x-7)

\int (3 x^{3} + 4 x - 7) d x

\frac{3 x ^{4}}{4} + 2 x^{2} - 7 x + C

Solution steps

\int 3 x^{3} + 4 x - 7 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 4 x d x - \int 7 d x

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

\int 4 x d x = 2 x^{2}

\int 7 d x = 7 x

= \frac{3 x ^{4}}{4} + 2 x^{2} - 7 x

Add a constant to the solution

= \frac{3 x ^{4}}{4} + 2 x^{2} - 7 x + C

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What is the integral of (3x^3+4x-7) ?
The integral of (3x^3+4x-7) is (3x^4)/4+2x^2-7x+C