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

The integral of x^2sqrt(2x-1) is 1/105 (15x^2+6x+2)(2x-1)^{3/2}+C

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

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

\frac{1}{105} (15 x^{2} + 6 x + 2) (2 x - 1)^{\frac{3}{2}} + C

Solution steps

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

Apply u-substitution

: \int \frac{u ( u + 1 ) ^{2}}{8} d u

= \int \frac{u ( u + 1 ) ^{2}}{8} d u

Take the constant out:

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

= \frac{1}{8} \cdot \int u (u + 1)^{2} d u

Expand

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

= \frac{1}{8} \cdot \int u^{\frac{5}{2}} + 2 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

= \frac{1}{8} (\int u^{\frac{5}{2}} d u + \int 2 u^{\frac{3}{2}} d u + \int u d u)

\int u^{\frac{5}{2}} d u = \frac{2}{7} u^{\frac{7}{2}}

\int 2 u^{\frac{3}{2}} d u = \frac{4}{5} u^{\frac{5}{2}}

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

= \frac{1}{8} (\frac{2}{7} u^{\frac{7}{2}} + \frac{4}{5} u^{\frac{5}{2}} + \frac{2}{3} u^{\frac{3}{2}})

Substitute back

u = 2 x - 1

= \frac{1}{8} (\frac{2}{7} (2 x - 1)^{\frac{7}{2}} + \frac{4}{5} (2 x - 1)^{\frac{5}{2}} + \frac{2}{3} (2 x - 1)^{\frac{3}{2}})

Simplify

\frac{1}{8} (\frac{2}{7} (2 x - 1)^{\frac{7}{2}} + \frac{4}{5} (2 x - 1)^{\frac{5}{2}} + \frac{2}{3} (2 x - 1)^{\frac{3}{2}}) : \frac{1}{105} (15 x^{2} + 6 x + 2) (2 x - 1)^{\frac{3}{2}}

= \frac{1}{105} (15 x^{2} + 6 x + 2) (2 x - 1)^{\frac{3}{2}}

Add a constant to the solution

= \frac{1}{105} (15 x^{2} + 6 x + 2) (2 x - 1)^{\frac{3}{2}} + C

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What is the integral of x^2sqrt(2x-1) ?
The integral of x^2sqrt(2x-1) is 1/105 (15x^2+6x+2)(2x-1)^{3/2}+C