What is the integral of xsqrt(3+2x) ?

The integral of xsqrt(3+2x) is 1/5 (x-1)(2x+3)^{3/2}+C

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integral of xsqrt(3+2x)

\int x 3 + 2 x d x

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

Solution steps

\int x 3 + 2 x d x

Apply u-substitution

: \int \frac{u ( u - 3 )}{4} d u

= \int \frac{u ( u - 3 )}{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 - 3) d u

Expand

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

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

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

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

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

Substitute back

u = 3 + 2 x

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

Simplify

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

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

Add a constant to the solution

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

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What is the integral of xsqrt(3+2x) ?
The integral of xsqrt(3+2x) is 1/5 (x-1)(2x+3)^{3/2}+C