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

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

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integral of (4x^3-3x^2+2x-5)

\int (4 x^{3} - 3 x^{2} + 2 x - 5) d x

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

Solution steps

\int 4 x^{3} - 3 x^{2} + 2 x - 5 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^{2} d x + \int 2 x d x - \int 5 d x

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

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

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

\int 5 d x = 5 x

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

Add a constant to the solution

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

\frac{d}{d x} (f (\frac{1}{3} x^{3})) = 2 x^{5}

\frac{d}{d x} (2 cos (x) + (cos (x))^{2})

y^{^{'}} = \frac{y ^{4} x ^{4} + 4 x ^{4}}{y ^{3}}

\int \frac{x + 7}{x ^{2} ( x + 2 )} d x

\frac{d}{d x} (\frac{cos ( x ) 2 x ^{3}}{e ^{x}})

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