What is the integral of (x^3+10x-3) ?

The integral of (x^3+10x-3) is (x^4)/4+5x^2-3x+C

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integral of (x^3+10x-3)

\int (x^{3} + 10 x - 3) d x

\frac{x ^{4}}{4} + 5 x^{2} - 3 x + C

Solution steps

\int x^{3} + 10 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 10 x d x - \int 3 d x

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

\int 10 x d x = 5 x^{2}

\int 3 d x = 3 x

= \frac{x ^{4}}{4} + 5 x^{2} - 3 x

Add a constant to the solution

= \frac{x ^{4}}{4} + 5 x^{2} - 3 x + C

\frac{8}{s ( s + 8 )}

\frac{\partial}{\partial x} (n ln (\frac{x}{2}))

g (t) = \frac{7}{8 t ^{4}}

\int sin^{4} (x) cos^{4} (x) d x

\frac{\partial}{\partial x} (z (1 + e^{x + y}))

What is the integral of (x^3+10x-3) ?
The integral of (x^3+10x-3) is (x^4)/4+5x^2-3x+C