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

The integral of (x^2)/(sqrt(2x-1)) is 1/15 (3x^2+2x+2)sqrt(2x-1)+C

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

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

\frac{1}{15} (3 x^{2} + 2 x + 2) 2 x - 1 + C

Solution steps

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

Apply u-substitution

: \int \frac{( u ^{2} + 1 ) ^{2}}{4} d u

= \int \frac{( u ^{2} + 1 ) ^{2}}{4} d u

Take the constant out:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

= \frac{1}{4} \cdot \int (u^{2} + 1)^{2} d u

Expand

(u^{2} + 1)^{2} : u^{4} + 2 u^{2} + 1

= \frac{1}{4} \cdot \int u^{4} + 2 u^{2} + 1 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

= \frac{1}{4} (\int u^{4} d u + \int 2 u^{2} d u + \int 1 d u)

\int u^{4} d u = \frac{u ^{5}}{5}

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

\int 1 d u = u

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

Substitute back

u = 2 x - 1

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

Simplify

\frac{1}{4} (\frac{( 2 x - 1 ) ^{5}}{5} + \frac{2 ( 2 x - 1 ) ^{3}}{3} + 2 x - 1) : \frac{1}{15} (3 x^{2} + 2 x + 2) 2 x - 1

= \frac{1}{15} (3 x^{2} + 2 x + 2) 2 x - 1

Add a constant to the solution

= \frac{1}{15} (3 x^{2} + 2 x + 2) 2 x - 1 + C

\frac{8}{( 9 s ^{2} + 12 s + 4 ) ( s )}

\int x \cdot e^{x^{2}} d x

\frac{d ^{2} x}{d t ^{2}} + 9 x = - sin (3 t)

\frac{\partial}{\partial x} (ln (\frac{2 x y}{z}))

\int \frac{5 x}{3 x ^{\frac{1}{3}}} d x

What is the integral of (x^2)/(sqrt(2x-1)) ?
The integral of (x^2)/(sqrt(2x-1)) is 1/15 (3x^2+2x+2)sqrt(2x-1)+C