What is the integral of (x-3)sqrt(2x+1) ?

The integral of (x-3)sqrt(2x+1) is 1/5 x(2x+1)^{3/2}-16/15 (2x+1)^{3/2}+C

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integral of (x-3)sqrt(2x+1)

\int (x - 3) 2 x + 1 d x

\frac{1}{5} x (2 x + 1)^{\frac{3}{2}} - \frac{16}{15} (2 x + 1)^{\frac{3}{2}} + C

Solution steps

\int (x - 3) 2 x + 1 d x

Expand

(x - 3) 2 x + 1 : x 2 x + 1 - 3 2 x + 1

= \int x 2 x + 1 - 3 2 x + 1 d x

Apply the Sum Rule:

\int f (x) \pm g (x) d x = \int f (x) d x \pm \int g (x) d x

= \int x 2 x + 1 d x - \int 3 2 x + 1 d x

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

\int 3 2 x + 1 d x = (2 x + 1)^{\frac{3}{2}}

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

Simplify

\frac{1}{15} (3 x - 1) (2 x + 1)^{\frac{3}{2}} - (2 x + 1)^{\frac{3}{2}} : \frac{1}{5} x (2 x + 1)^{\frac{3}{2}} - \frac{16}{15} (2 x + 1)^{\frac{3}{2}}

= \frac{1}{5} x (2 x + 1)^{\frac{3}{2}} - \frac{16}{15} (2 x + 1)^{\frac{3}{2}}

Add a constant to the solution

= \frac{1}{5} x (2 x + 1)^{\frac{3}{2}} - \frac{16}{15} (2 x + 1)^{\frac{3}{2}} + C

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What is the integral of (x-3)sqrt(2x+1) ?
The integral of (x-3)sqrt(2x+1) is 1/5 x(2x+1)^{3/2}-16/15 (2x+1)^{3/2}+C