What is the integral of sqrt(1+(x-1)^{1/2)} ?

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

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integral of sqrt(1+(x-1)^{1/2)}

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

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

Solution steps

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

Apply u-substitution

: \int 1 + u \cdot 2 u d u

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

Take the constant out:

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

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

Apply u-substitution

: \int (v - 1) v d v

= 2 \cdot \int (v - 1) v d v

Expand

(v - 1) v : v^{\frac{3}{2}} - v

= 2 \cdot \int v^{\frac{3}{2}} - v d v

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 v^{\frac{3}{2}} d v - \int v d v)

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

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

= 2 (\frac{2}{5} v^{\frac{5}{2}} - \frac{2}{3} v^{\frac{3}{2}})

Substitute back

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

Apply radical rule:

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

Add a constant to the solution

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

(- 1, 6), (2, - 3)

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