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

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

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integral of xsqrt(1+2x)

\int x 1 + 2 x d x

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

Solution steps

\int x 1 + 2 x d x

Apply u-substitution

= \int \frac{u ( u - 1 )}{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 (u - 1) d u

Expand

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

= \frac{1}{4} \cdot \int 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}{4} (\int u^{\frac{3}{2}} d u - \int u d u)

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

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

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

Substitute back

u = 1 + 2 x

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

Simplify

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

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

Add a constant to the solution

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

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