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

The integral of tsqrt(t-1) is 2/5 (t-1)^{5/2}+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 tsqrt(t-1)

\int t t - 1 d t

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

Solution steps

\int t t - 1 d t

Apply u-substitution

: \int (u + 1) u d u

= \int (u + 1) u d u

Expand

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

= \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

= \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}}

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

Substitute back

u = t - 1

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

Add a constant to the solution

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

\int \frac{1}{x 4 x ^{2} + 1} d x

(cosh (x))^{^{'}}

\frac{d}{d x} (25)

x ln (x^{2} + 2)

\frac{d}{d y} ((y - 1)^{2})

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