What is the integral of (10x+1)sqrt(4x-1) ?

The integral of (10x+1)sqrt(4x-1) is x(4x-1)^{3/2}+1/3 (4x-1)^{3/2}+C

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integral of (10x+1)sqrt(4x-1)

\int (10 x + 1) 4 x - 1 d x

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

Solution steps

\int (10 x + 1) 4 x - 1 d x

Expand

(10 x + 1) 4 x - 1 : 10 x 4 x - 1 + 4 x - 1

= \int 10 x 4 x - 1 + 4 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 10 x 4 x - 1 d x + \int 4 x - 1 d x

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

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

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

Simplify

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

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

Add a constant to the solution

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

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What is the integral of (10x+1)sqrt(4x-1) ?
The integral of (10x+1)sqrt(4x-1) is x(4x-1)^{3/2}+1/3 (4x-1)^{3/2}+C