What is the integral of (3x^8-7x^3+8) ?

The integral of (3x^8-7x^3+8) is (x^9)/3-(7x^4)/4+8x+C

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integral of (3x^8-7x^3+8)

\int (3 x^{8} - 7 x^{3} + 8) d x

\frac{x ^{9}}{3} - \frac{7 x ^{4}}{4} + 8 x + C

Solution steps

\int 3 x^{8} - 7 x^{3} + 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 3 x^{8} d x - \int 7 x^{3} d x + \int 8 d x

\int 3 x^{8} d x = \frac{x ^{9}}{3}

\int 7 x^{3} d x = \frac{7 x ^{4}}{4}

\int 8 d x = 8 x

= \frac{x ^{9}}{3} - \frac{7 x ^{4}}{4} + 8 x

Add a constant to the solution

= \frac{x ^{9}}{3} - \frac{7 x ^{4}}{4} + 8 x + C

What is the integral of (3x^8-7x^3+8) ?
The integral of (3x^8-7x^3+8) is (x^9)/3-(7x^4)/4+8x+C