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

The integral of (x^2)/(sqrt(x-1)) is 2/5 (x-1)^{5/2}+4/3 (x-1)^{3/2}+2sqrt(x-1)+C

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

integral of (x^2)/(sqrt(x-1))

integral

\frac{x ^{2}}{x - 1}

\frac{2}{5} (x - 1)^{\frac{5}{2}} + \frac{4}{3} (x - 1)^{\frac{3}{2}} + 2 x - 1 + C

Solution steps

\int \frac{x ^{2}}{x - 1} d x

Apply u-substitution

= \int \frac{( u + 1 ) ^{2}}{u} d u

Expand

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

= \int u^{\frac{3}{2}} + 2 u + \frac{1}{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 2 u d u + \int \frac{1}{u} d u

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

\int 2 u d u = \frac{4}{3} u^{\frac{3}{2}}

\int \frac{1}{u} d u = 2 u

= \frac{2}{5} u^{\frac{5}{2}} + \frac{4}{3} u^{\frac{3}{2}} + 2 u

Substitute back

u = x - 1

= \frac{2}{5} (x - 1)^{\frac{5}{2}} + \frac{4}{3} (x - 1)^{\frac{3}{2}} + 2 x - 1

Add a constant to the solution

= \frac{2}{5} (x - 1)^{\frac{5}{2}} + \frac{4}{3} (x - 1)^{\frac{3}{2}} + 2 x - 1 + C

What is the integral of (x^2)/(sqrt(x-1)) ?
The integral of (x^2)/(sqrt(x-1)) is 2/5 (x-1)^{5/2}+4/3 (x-1)^{3/2}+2sqrt(x-1)+C