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

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

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integral of sqrt(x-1)

\int x - 1 d x

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

Solution steps

\int x - 1 d x

Apply u-substitution

: \int u d u

= \int u d u

Apply the Power Rule

: \frac{2}{3} u^{\frac{3}{2}}

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

Substitute back

u = x - 1

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

Add a constant to the solution

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

\int_{1}^{2} x x - 1 d x

\frac{1}{s ^{2} ( s + \frac{1}{15} )}

\frac{d}{d x} (a x^{4})

n = 1 \sum \infty n^{2} (\frac{5}{6})^{n}

\int (1 - 3 y)^{2} - (1 - y^{2})^{2} d y

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