Question

\int (x+1)\sqrt(2x-1)dx

Answer 1

`\int (x+ 1) √(2x-1) dx = (1)/(3)(x+1) (2x - 1)^{(3)/(2) } - \ (1)/(15)(2x-1)^{(5)/(2)} + C`

Explanation:

`\int (x+1)\sqrt {(2x-1)} dx\\Integrate \ using \ integration \ by\ parts \\\\u = x + 1, v'= √(2x - 1)\\\\v'= √(2x - 1)\\\\integrate \ both \ sides \\\\\int v'= \int √(2x- 1)dx\\\\v = \int ( 2x - 1)^{(1)/(2) } \ dx\\\\v = \frac{(2x - 1)^{(1)/(2) + 1}}{(1)/(2) + 1}} * (1)/(2)\\\\v= \frac{(2x - 1)^{(3)/(2)}}{(3)/(2)} * (1)/(2)\\\\v = \frac{2 * (2x - 1)^{(3)/(2)}}{3} * (1)/(2)\\\\v = \frac{(2x - 1)^{(3)/(2)}}{3}`

`\int (x+1)\sqrt(2x-1)dx\\\\ = uv - \int v du`

`= (x +1 ) \cdot \frac{(2x - 1)^{(3)/(2)}}{3} - \int \frac{(2x - 1)^{(3)/(2)}}{3} dx \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ [ \ u = x + 1 => du = dx \ ]`

`= (1)/(3)(x+1) (2x - 1)^{(3)/(2) } - \ (1)/(3) \int (2x - 1)^{(3)/(2)}} dx\\\\= (1)/(3)(x+1) (2x - 1)^{(3)/(2) } - \ (1)/(3) * ( \frac{(2x-1)^{(3)/(2) + 1}}{(3)/(2) + 1}) * (1)/(2)\\\\= (1)/(3)(x+1) (2x - 1)^{(3)/(2) } - \ (1)/(3) * ( \frac{(2x-1)^{(5)/(2)}}{(5)/(2) }) * (1)/(2)\\\\= (1)/(3)(x+1) (2x - 1)^{(3)/(2) } - \ (1)/(15) * (2x-1)^{(5)/(2)} + C\\\\`