Question

Evaluate ∫xlog(x + 3)dx.​

Answer 1

Answer:

$∫x log(x + 3)$

Important formula:

$∫udv = uv - ∫vdu$

Solution:

$let \: u = log(x + 3) \\ = > du = \frac{1}{x + 3}$

$dv = x$

$= > v = ∫x$

$= > v = \frac{ {x}^{2} }{2}$

To find:

$∫ log(x + 3) x$

$uv = \frac{ {x}^{2} }{2} log(x + 3)$

$vdu = \frac{ {x}^{2} }{2} \times \frac{1}{x + 3} = \frac{ {x}^{2} }{2(x + 3)}$

$∫vdu = ∫ \frac{ {x}^{2} }{2(x + 3)}$

$= ∫ \frac{ {x}^{2} - 9 + 9}{2(x + 3)}$

$= \frac{1}{2} ∫ \frac{ {x}^{2} - 9 }{x + 3} + \frac{1}{2} ∫ \frac{9}{x + 3}$

$= \frac{1}{2} ∫ \frac{(x + 3)(x - 3)}{x + 3} + \frac{9}{2} ∫ \frac{1}{x + 3}$

$=( \frac{1}{2} ∫x - 3) + \frac{9}{2} log(x + 3)$

$= \frac{1}{2} ( \frac{ {x}^{2} }{2} - 3x) + \frac{9}{2} log(x + 3)$

$uv - ∫vdu$

$= \frac{ {x}^{2} }{2} log(x + 3) - \frac{1}{2} ( \frac{ {x}^{2} }{2} - 3x) - \frac{9}{2} log(x + 3)$

$= \frac{ {x}^{2} - 9 }{2} log(x + 3) - \frac{1}{2} ( \frac{ {x}^{2} }{2} - 3x)$