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

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

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

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

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

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

Solution steps

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

Expand

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

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

Apply the Sum Rule:

\int f (x) \pm g (x) d x = \int f (x) d x \pm \int g (x) d x

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

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

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

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

Simplify

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

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

Add a constant to the solution

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

\int x^{3} sin (4 x^{4} - π) d x

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

\int (x^{2} + 2) ln (3 x) d x

\int sin^{3} (x) cos (cos (x)) d x

\int 3 sin^{5} (x) cos^{4} (x) d x

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