What is the integral of x^3sqrt(x-1) ?

The integral of x^3sqrt(x-1) is 2/9 (x-1)^{9/2}+6/7 (x-1)^{7/2}+6/5 (x-1)^{5/2}+2/3 (x-1)^{3/2}+C

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integral of x^3sqrt(x-1)

\int x^{3} x - 1 d x

\frac{2}{9} (x - 1)^{\frac{9}{2}} + \frac{6}{7} (x - 1)^{\frac{7}{2}} + \frac{6}{5} (x - 1)^{\frac{5}{2}} + \frac{2}{3} (x - 1)^{\frac{3}{2}} + C

Solution steps

\int x^{3} x - 1 d x

Apply u-substitution

: \int (u + 1)^{3} u d u

= \int (u + 1)^{3} u d u

Expand

(u + 1)^{3} u : u^{\frac{7}{2}} + 3 u^{\frac{5}{2}} + 3 u^{\frac{3}{2}} + u

= \int u^{\frac{7}{2}} + 3 u^{\frac{5}{2}} + 3 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{7}{2}} d u + \int 3 u^{\frac{5}{2}} d u + \int 3 u^{\frac{3}{2}} d u + \int u d u

\int u^{\frac{7}{2}} d u = \frac{2}{9} u^{\frac{9}{2}}

\int 3 u^{\frac{5}{2}} d u = \frac{6}{7} u^{\frac{7}{2}}

\int 3 u^{\frac{3}{2}} d u = \frac{6}{5} u^{\frac{5}{2}}

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

= \frac{2}{9} u^{\frac{9}{2}} + \frac{6}{7} u^{\frac{7}{2}} + \frac{6}{5} u^{\frac{5}{2}} + \frac{2}{3} u^{\frac{3}{2}}

Substitute back

u = x - 1

= \frac{2}{9} (x - 1)^{\frac{9}{2}} + \frac{6}{7} (x - 1)^{\frac{7}{2}} + \frac{6}{5} (x - 1)^{\frac{5}{2}} + \frac{2}{3} (x - 1)^{\frac{3}{2}}

Add a constant to the solution

= \frac{2}{9} (x - 1)^{\frac{9}{2}} + \frac{6}{7} (x - 1)^{\frac{7}{2}} + \frac{6}{5} (x - 1)^{\frac{5}{2}} + \frac{2}{3} (x - 1)^{\frac{3}{2}} + C

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What is the integral of x^3sqrt(x-1) ?
The integral of x^3sqrt(x-1) is 2/9 (x-1)^{9/2}+6/7 (x-1)^{7/2}+6/5 (x-1)^{5/2}+2/3 (x-1)^{3/2}+C