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

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

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

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

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

Solution steps

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

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

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

\int 3 d x = 3 x

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

Add a constant to the solution

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

5 x^{- 15}

\int \frac{10 x + 15}{x ^{2} + 3 x} d x

\frac{d y}{d x} = \frac{2 - x}{y - 3}

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

\frac{\partial}{\partial v} (\frac{v}{u})

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