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

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

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

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

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

Solution steps

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

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

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

\int 5 d x = 5 x

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

Add a constant to the solution

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

\frac{d y}{d x} = \frac{x + y ^{2}}{2 y}

y = 2 x^{3} - 8 x

\int \frac{e ^{z} + 1}{e ^{z} + z} d z

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

t \to 0 lim (\frac{4}{t} - \frac{4}{t ^{2} + t})

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