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

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

\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 + C

Solution steps

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

Apply u-substitution

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

= \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

x \to - 5 - lim (3 x + 15)

\frac{d}{d a} ((3 a^{2})^{3})

(x^{3} + 8 x^{3} y^{2}) d x + e^{x^{4}} y d y = 0

\int a + b x d x

x \to 0 lim (\frac{1 - cos ( 2 x )}{x})

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