What is the integral of (x^3-2x^{1/2}+4)^2 ?

The integral of (x^3-2x^{1/2}+4)^2 is (x^7)/7-8/9 x^{9/2}+2x^4+2x^2-32/3 x^{3/2}+16x+C

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integral of (x^3-2x^{1/2}+4)^2

\int (x^{3} - 2 x^{\frac{1}{2}} + 4)^{2} d x

\frac{x ^{7}}{7} - \frac{8}{9} x^{\frac{9}{2}} + 2 x^{4} + 2 x^{2} - \frac{32}{3} x^{\frac{3}{2}} + 16 x + C

Solution steps

\int (x^{3} - 2 x^{\frac{1}{2}} + 4)^{2} d x

Simplify

(x^{3} - 2 x^{\frac{1}{2}} + 4)^{2} : x^{6} - 4 x^{\frac{7}{2}} + 8 x^{3} + 4 x - 16 x^{\frac{1}{2}} + 16

= \int x^{6} - 4 x^{\frac{7}{2}} + 8 x^{3} + 4 x - 16 x^{\frac{1}{2}} + 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^{6} d x - \int 4 x^{\frac{7}{2}} d x + \int 8 x^{3} d x + \int 4 x d x - \int 16 x^{\frac{1}{2}} d x + \int 16 d x

\int x^{6} d x = \frac{x ^{7}}{7}

\int 4 x^{\frac{7}{2}} d x = \frac{8}{9} x^{\frac{9}{2}}

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

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

\int 16 x^{\frac{1}{2}} d x = \frac{32}{3} x^{\frac{3}{2}}

\int 16 d x = 16 x

= \frac{x ^{7}}{7} - \frac{8}{9} x^{\frac{9}{2}} + 2 x^{4} + 2 x^{2} - \frac{32}{3} x^{\frac{3}{2}} + 16 x

Add a constant to the solution

= \frac{x ^{7}}{7} - \frac{8}{9} x^{\frac{9}{2}} + 2 x^{4} + 2 x^{2} - \frac{32}{3} x^{\frac{3}{2}} + 16 x + C

\frac{d}{d x} (\frac{1}{y - 5 x ^{2}})

y^{^{'}} - y = 2 t

x^{2} y^{^{'}} + 2 x y = 25 cos^{2} (x)

\frac{- 8 x}{x ^{2} + 1}

\frac{5}{s ( s ^{2} + 4 s + 5 )}

What is the integral of (x^3-2x^{1/2}+4)^2 ?
The integral of (x^3-2x^{1/2}+4)^2 is (x^7)/7-8/9 x^{9/2}+2x^4+2x^2-32/3 x^{3/2}+16x+C