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

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

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integral of (x+1)sqrt(x-1)

\int (x + 1) x - 1 d x

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

Solution steps

\int (x + 1) x - 1 d x

Apply u-substitution

: \int u (u + 2) d u

= \int u (u + 2) d u

Expand

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

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

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

\int 2 u d u = \frac{4}{3} u^{\frac{3}{2}}

= \frac{2}{5} u^{\frac{5}{2}} + \frac{4}{3} u^{\frac{3}{2}}

Substitute back

u = x - 1

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

Add a constant to the solution

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

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