What is the integral of (4x^3+3x-1) ?

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

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integral of (4x^3+3x-1)

\int (4 x^{3} + 3 x - 1) d x

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

Solution steps

\int 4 x^{3} + 3 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 4 x^{3} d x + \int 3 x d x - \int 1 d x

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

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

\int 1 d x = x

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

Add a constant to the solution

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

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