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

The integral of (x) 3/4 x^2-4x+8 is (3x^4)/(16)-2x^2+8x+C

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

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

\frac{3 x ^{4}}{16} - 2 x^{2} + 8 x + C

Solution steps

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

\int x \frac{3}{4} x^{2} d x = \frac{3 x ^{4}}{16}

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

\int 8 d x = 8 x

= \frac{3 x ^{4}}{16} - 2 x^{2} + 8 x

Add a constant to the solution

= \frac{3 x ^{4}}{16} - 2 x^{2} + 8 x + C

What is the integral of (x) 3/4 x^2-4x+8 ?
The integral of (x) 3/4 x^2-4x+8 is (3x^4)/(16)-2x^2+8x+C