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

\frac{d}{d x} (sin^{4} (x) + cos^{4} (x))

2 cos (2 t)

y^{^{''}} + 8 y^{^{'}} + 25 y = 0, y (3319) = - 3, y^{^{'}} (3319) = 1

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

\frac{d}{d x} (x^{2} (20 - x)^{3})

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