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

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

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

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

= \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 = 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

\int e^{x} (y + 2 z) d z

\int \frac{3 ^{2 + x} - 5}{x} d x

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

\int \frac{3 x ^{2} + 2 x}{x + 1} d x

\int \frac{5}{( 5 x - 1 ) ln ( 5 x - 1 )} d x

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