What is the integral of (6x+1)*sqrt(2x-3) ?

The integral of (6x+1)*sqrt(2x-3) is 6/5 x(2x-3)^{3/2}+23/15 (2x-3)^{3/2}+C

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integral of (6x+1)*sqrt(2x-3)

\int (6 x + 1) \cdot 2 x - 3 d x

\frac{6}{5} x (2 x - 3)^{\frac{3}{2}} + \frac{23}{15} (2 x - 3)^{\frac{3}{2}} + C

Solution steps

\int (6 x + 1) 2 x - 3 d x

Expand

(6 x + 1) 2 x - 3 : 6 x 2 x - 3 + 2 x - 3

= \int 6 x 2 x - 3 + 2 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 6 x 2 x - 3 d x + \int 2 x - 3 d x

\int 6 x 2 x - 3 d x = \frac{6}{5} (x + 1) (2 x - 3)^{\frac{3}{2}}

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

= \frac{6}{5} (x + 1) (2 x - 3)^{\frac{3}{2}} + \frac{1}{3} (2 x - 3)^{\frac{3}{2}}

Simplify

\frac{6}{5} (x + 1) (2 x - 3)^{\frac{3}{2}} + \frac{1}{3} (2 x - 3)^{\frac{3}{2}} : \frac{6}{5} x (2 x - 3)^{\frac{3}{2}} + \frac{23}{15} (2 x - 3)^{\frac{3}{2}}

= \frac{6}{5} x (2 x - 3)^{\frac{3}{2}} + \frac{23}{15} (2 x - 3)^{\frac{3}{2}}

Add a constant to the solution

= \frac{6}{5} x (2 x - 3)^{\frac{3}{2}} + \frac{23}{15} (2 x - 3)^{\frac{3}{2}} + C

f (x) = (1 + 2 x) (3 - 2 x), at x = 1

\int x^{\frac{3}{5}} d x

\int_{2.4}^{0.9} \frac{5}{6.4 - 10 x} d x

\frac{x ^{2}}{sin ( x )}

x \to - 1 lim (\frac{x ^{5} + 1}{x + 1})

What is the integral of (6x+1)*sqrt(2x-3) ?
The integral of (6x+1)*sqrt(2x-3) is 6/5 x(2x-3)^{3/2}+23/15 (2x-3)^{3/2}+C