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

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

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

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

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

Solution steps

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

Apply u-substitution

= \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 = 1 + 2 x

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

Simplify

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

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