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

The integral of sqrt(x)(4(x-1)^2+1/(x^3)) is -2/(3x^{3/2)}+8/7 x^{7/2}-16/5 x^{5/2}+8/3 x^{3/2}+C

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

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

- \frac{2}{3 x ^{\frac{3}{2}}} + \frac{8}{7} x^{\frac{7}{2}} - \frac{16}{5} x^{\frac{5}{2}} + \frac{8}{3} x^{\frac{3}{2}} + C

Solution steps

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

Expand

x (4 (x - 1)^{2} + \frac{1}{x ^{3}}) : \frac{1}{x ^{\frac{5}{2}}} + 4 x^{\frac{5}{2}} - 8 x^{\frac{3}{2}} + 4 x^{\frac{1}{2}}

= \int \frac{1}{x ^{\frac{5}{2}}} + 4 x^{\frac{5}{2}} - 8 x^{\frac{3}{2}} + 4 x^{\frac{1}{2}} 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 \frac{1}{x ^{\frac{5}{2}}} d x + \int 4 x^{\frac{5}{2}} d x - \int 8 x^{\frac{3}{2}} d x + \int 4 x^{\frac{1}{2}} d x

\int \frac{1}{x ^{\frac{5}{2}}} d x = - \frac{2}{3 x ^{\frac{3}{2}}}

\int 4 x^{\frac{5}{2}} d x = \frac{8}{7} x^{\frac{7}{2}}

\int 8 x^{\frac{3}{2}} d x = \frac{16}{5} x^{\frac{5}{2}}

\int 4 x^{\frac{1}{2}} d x = \frac{8}{3} x^{\frac{3}{2}}

= - \frac{2}{3 x ^{\frac{3}{2}}} + \frac{8}{7} x^{\frac{7}{2}} - \frac{16}{5} x^{\frac{5}{2}} + \frac{8}{3} x^{\frac{3}{2}}

Add a constant to the solution

= - \frac{2}{3 x ^{\frac{3}{2}}} + \frac{8}{7} x^{\frac{7}{2}} - \frac{16}{5} x^{\frac{5}{2}} + \frac{8}{3} x^{\frac{3}{2}} + C

What is the integral of sqrt(x)(4(x-1)^2+1/(x^3)) ?
The integral of sqrt(x)(4(x-1)^2+1/(x^3)) is -2/(3x^{3/2)}+8/7 x^{7/2}-16/5 x^{5/2}+8/3 x^{3/2}+C