What is the integral of x(sqrt(x-1)-(x-1)^2) ?

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

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integral of x(sqrt(x-1)-(x-1)^2)

\int x (x - 1 - (x - 1)^{2}) d x

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

Solution steps

\int x (x - 1 - (x - 1)^{2}) d x

Expand

x (x - 1 - (x - 1)^{2}) : x x - 1 - x^{3} + 2 x^{2} - x

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

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

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

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

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

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

Add a constant to the solution

= \frac{2}{5} (x - 1)^{\frac{5}{2}} + \frac{2}{3} (x - 1)^{\frac{3}{2}} - \frac{x ^{4}}{4} + \frac{2 x ^{3}}{3} - \frac{x ^{2}}{2} + C

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