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

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

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

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

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

Solution steps

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

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

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

\int 4 d x = 4 x

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

Add a constant to the solution

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

\frac{d y}{d x} = \frac{1}{x} (y^{2} + 0.2 y), y (1) = 1.3

\int - 7 sec^{2} (x) tan (x) d x

\frac{d}{d x} (\frac{5}{x + 1} - x^{4})

n = 1 \sum \infty n (x - 3)^{n}

\int x^{3} 9 x^{4} + 1 d x

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