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

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

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

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

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

Solution steps

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

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

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

\int 5 d x = 5 x

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

Add a constant to the solution

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

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

x \to 1 lim (\frac{x + 2}{x - 1})

e^{\frac{t}{3}}

t sin (0.32 t)

\int \frac{1}{x 9 x ^{2} + 4} d x

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