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

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

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integral of (2x^3+10x^2-3x+4)

\int (2 x^{3} + 10 x^{2} - 3 x + 4) d x

\frac{x ^{4}}{2} + \frac{10 x ^{3}}{3} - \frac{3 x ^{2}}{2} + 4 x + C

Solution steps

\int 2 x^{3} + 10 x^{2} - 3 x + 4 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 2 x^{3} d x + \int 10 x^{2} d x - \int 3 x d x + \int 4 d x

\int 2 x^{3} d x = \frac{x ^{4}}{2}

\int 10 x^{2} d x = \frac{10 x ^{3}}{3}

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

\int 4 d x = 4 x

= \frac{x ^{4}}{2} + \frac{10 x ^{3}}{3} - \frac{3 x ^{2}}{2} + 4 x

Add a constant to the solution

= \frac{x ^{4}}{2} + \frac{10 x ^{3}}{3} - \frac{3 x ^{2}}{2} + 4 x + C

y^{^{''}} - 6 y^{^{'}} = 4 t

\frac{d}{d x} (\frac{2 x + 1}{3 x - 1})

\int (x - 1) (x + 2) (x + 4) d x

\frac{d y}{d x} = \frac{( y - x ^{2} + y ^{2} )}{x}

y y^{^{'}} = (y^{2} - y) (2 x + 5)

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