What is the integral of (4x^3+x^2-8x+12) ?

The integral of (4x^3+x^2-8x+12) is x^4+(x^3)/3-4x^2+12x+C

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integral of (4x^3+x^2-8x+12)

\int (4 x^{3} + x^{2} - 8 x + 12) d x

x^{4} + \frac{x ^{3}}{3} - 4 x^{2} + 12 x + C

Solution steps

\int 4 x^{3} + x^{2} - 8 x + 12 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 x^{2} d x - \int 8 x d x + \int 12 d x

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

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

\int 8 x d x = 4 x^{2}

\int 12 d x = 12 x

= x^{4} + \frac{x ^{3}}{3} - 4 x^{2} + 12 x

Add a constant to the solution

= x^{4} + \frac{x ^{3}}{3} - 4 x^{2} + 12 x + C

\int \frac{1}{2 u ^{\frac{1}{2}}} d u

\int (x^{2} - 6 x + 8) (x - 3) d x

f (x) = \frac{5}{x}

\int \frac{4 ^{x + 2}}{5 ^{x - 1}} d x

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

What is the integral of (4x^3+x^2-8x+12) ?
The integral of (4x^3+x^2-8x+12) is x^4+(x^3)/3-4x^2+12x+C