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

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

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integral of sqrt(t-1)

\int t - 1 d t

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

Solution steps

\int t - 1 d t

Apply u-substitution

= \int u d u

Apply the Power Rule

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

Substitute back

u = t - 1

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

Add a constant to the solution

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

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