What is the integral of (2x+1)sqrt(4x-3) ?

The integral of (2x+1)sqrt(4x-3) is 1/5 x(4x-3)^{3/2}+4/15 (4x-3)^{3/2}+C

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integral of (2x+1)sqrt(4x-3)

\int (2 x + 1) 4 x - 3 d x

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

Solution steps

\int (2 x + 1) 4 x - 3 d x

Expand

(2 x + 1) 4 x - 3 : 2 x 4 x - 3 + 4 x - 3

= \int 2 x 4 x - 3 + 4 x - 3 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 2 x 4 x - 3 d x + \int 4 x - 3 d x

\int 2 x 4 x - 3 d x = \frac{1}{10} (2 x + 1) (4 x - 3)^{\frac{3}{2}}

\int 4 x - 3 d x = \frac{1}{6} (4 x - 3)^{\frac{3}{2}}

= \frac{1}{10} (2 x + 1) (4 x - 3)^{\frac{3}{2}} + \frac{1}{6} (4 x - 3)^{\frac{3}{2}}

Simplify

\frac{1}{10} (2 x + 1) (4 x - 3)^{\frac{3}{2}} + \frac{1}{6} (4 x - 3)^{\frac{3}{2}} : \frac{1}{5} x (4 x - 3)^{\frac{3}{2}} + \frac{4}{15} (4 x - 3)^{\frac{3}{2}}

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

Add a constant to the solution

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

What is the integral of (2x+1)sqrt(4x-3) ?
The integral of (2x+1)sqrt(4x-3) is 1/5 x(4x-3)^{3/2}+4/15 (4x-3)^{3/2}+C