What is the integral of (x^2+8)(sqrt(x)-8) ?

The integral of (x^2+8)(sqrt(x)-8) is 2/7 x^{7/2}-(8x^3)/3+16/3 x^{3/2}-64x+C

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integral of (x^2+8)(sqrt(x)-8)

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

\frac{2}{7} x^{\frac{7}{2}} - \frac{8 x ^{3}}{3} + \frac{16}{3} x^{\frac{3}{2}} - 64 x + C

Solution steps

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

Expand

(x^{2} + 8) (x - 8) : x^{\frac{5}{2}} - 8 x^{2} + 8 x - 64

= \int x^{\frac{5}{2}} - 8 x^{2} + 8 x - 64 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{5}{2}} d x - \int 8 x^{2} d x + \int 8 x d x - \int 64 d x

\int x^{\frac{5}{2}} d x = \frac{2}{7} x^{\frac{7}{2}}

\int 8 x^{2} d x = \frac{8 x ^{3}}{3}

\int 8 x d x = \frac{16}{3} x^{\frac{3}{2}}

\int 64 d x = 64 x

= \frac{2}{7} x^{\frac{7}{2}} - \frac{8 x ^{3}}{3} + \frac{16}{3} x^{\frac{3}{2}} - 64 x

Add a constant to the solution

= \frac{2}{7} x^{\frac{7}{2}} - \frac{8 x ^{3}}{3} + \frac{16}{3} x^{\frac{3}{2}} - 64 x + C

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What is the integral of (x^2+8)(sqrt(x)-8) ?
The integral of (x^2+8)(sqrt(x)-8) is 2/7 x^{7/2}-(8x^3)/3+16/3 x^{3/2}-64x+C