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

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

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integral of 2xsqrt(x-1)

\int 2 x x - 1 d x

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

Solution steps

\int 2 x x - 1 d x

Take the constant out:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

= 2 \cdot \int x x - 1 d x

Apply u-substitution

: \int (u + 1) u d u

= 2 \cdot \int (u + 1) u d u

Expand

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

= 2 \cdot \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

= 2 (\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}}

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

Substitute back

u = x - 1

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

Add a constant to the solution

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

\frac{d}{d x} (x (3 - 2 x)^{5})

y^{^{''}} + 4 y = t^{2} e^{3 t} + 3

x y^{^{'}} = y^{2} + 1

\frac{d y}{d x} + 3 y = x + e^{- 2 x}

\frac{3}{s + 2}

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