What is the integral of [xsqrt((x-1))] ?

The integral of [xsqrt((x-1))] is 2/5 (x-1)^{5/2}+2/3 (x-1)^{3/2}+C

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integral of [xsqrt((x-1))]

\int [x (x - 1)] d x

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

Solution steps

\int x x - 1 d x

Apply u-substitution

: \int (u + 1) u d u

= \int (u + 1) u d u

Expand

(u + 1) u : u^{\frac{3}{2}} + u

= \int u^{\frac{3}{2}} + u d u

Apply the Sum Rule:

\int f (x) \pm g (x) d x = \int f (x) d x \pm \int g (x) d x

= \int u^{\frac{3}{2}} d u + \int u d u

\int u^{\frac{3}{2}} d u = \frac{2}{5} u^{\frac{5}{2}}

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

= \frac{2}{5} u^{\frac{5}{2}} + \frac{2}{3} u^{\frac{3}{2}}

Substitute back

u = x - 1

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

Add a constant to the solution

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

x \to 1 + lim (\frac{4}{x ^{3} - 1})

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

f (x) = ln (sin (2 x))

y = 5 x^{2}, y = 30 x - 45, [0, 3]

\frac{d y}{d t} + \frac{3}{t} y = 3 t^{2} y^{2}

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