What is the integral of (x^2+4)(sqrt(x)-4) ?

The integral of (x^2+4)(sqrt(x)-4) is 2/7 x^{7/2}-(4x^3)/3+8/3 x^{3/2}-16x+C

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integral of (x^2+4)(sqrt(x)-4)

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

\frac{2}{7} x^{\frac{7}{2}} - \frac{4 x ^{3}}{3} + \frac{8}{3} x^{\frac{3}{2}} - 16 x + C

Solution steps

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

Expand

(x^{2} + 4) (x - 4) : x^{\frac{5}{2}} - 4 x^{2} + 4 x - 16

= \int x^{\frac{5}{2}} - 4 x^{2} + 4 x - 16 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 4 x^{2} d x + \int 4 x d x - \int 16 d x

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

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

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

\int 16 d x = 16 x

= \frac{2}{7} x^{\frac{7}{2}} - \frac{4 x ^{3}}{3} + \frac{8}{3} x^{\frac{3}{2}} - 16 x

Add a constant to the solution

= \frac{2}{7} x^{\frac{7}{2}} - \frac{4 x ^{3}}{3} + \frac{8}{3} x^{\frac{3}{2}} - 16 x + C

What is the integral of (x^2+4)(sqrt(x)-4) ?
The integral of (x^2+4)(sqrt(x)-4) is 2/7 x^{7/2}-(4x^3)/3+8/3 x^{3/2}-16x+C