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

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

\frac{d ^{2}}{d x ^{2}} (\frac{6}{7 x ^{5}})

\frac{d y}{d x} + 3 (tan (3 x)) y = - 6 cos (3 x)

\frac{d}{d x} ((2 x^{3}) (4 x^{2}))

(0, 5), (- 1.5, 1)

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