Question

Integrate by parts:-
$x {(logx)}^{2}$

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Answer 1

$\huge\tt\red{\bold{\underline{\underline{❥Question᎓}}}}$Integrate by parts:-

$x {(logx)}^{2}$

$\huge\mathcal{Answer}$

Given:-

$x {(logx)}^{2}$

Here ,this identity is used :-

$\bold{\boxed{∫(1 \times 2)dx = 1∫2dx - ∫(\frac{d(1)}{dx} \times ∫2dx)dx}}$

$⟹\bold{∫x( {logx)}^{2} dx = ∫ {(logx)}^{2} x \times dx}$

$⟹ \bold{{(logx)}^{2} ∫xdx - ∫ \frac{d(logx)}{dx} ∫x \times dx}$

$⟹\bold{\frac{ {x}^{2} }{2} {(logx)}^{2} - 2 ∫ \frac{logx}{x} \times \frac{ {x}^{2} }{2} dx}$

$⟹ \bold{\frac{ {x}^{2} }{2} {(logx)}^{2} - ∫xlogxdx......(i)}$

$\bold{\red{ɪ1=∫xlogxdx</p><p>∫xlogxdx=∫(logx)xdx}}$

$⟹\bold{logx∫xdx - ∫( \frac{d(logx)}{dx} ∫xdx)dx}$

$⟹\bold{logx( \frac{ {x}^{2} }{2} ) - ∫ \frac{1}{x} \times \frac{ {x}^{2} }{2} dx}$

$⟹\bold{ \frac{ {x}^{2} }{2} logx - \frac{1}{2} ∫xdx}$

$⟹ \bold{\frac{ {x}^{2} }{2} logx - \frac{1}{2} \frac{ {x}^{2} }{2} + c}$

$⟹\bold{ \frac{ {x}^{2} }{2} logx - \frac{ {x}^{2} }{4} + c}$

$\mathbb{\bold{Now,\:put\: the \:value \:of \:ɪ 1}}$

$⟹\bold{∫x {(logx)}^{2} dx = \frac{ {x}^{2} }{2} {(logx)}^{2} - ∫x \times logxdx}$

$⟹\bold{ \frac{ {x}^{2} }{2} {(logx)}^{2} - (\frac{ {x}^{2} (logx)}{2} - \frac{ {x}^{2} }{4} + c)}$

$⟹ \bold{\frac{ {x}^{2} }{2} {(logx)}^{2} - \frac{ {x}^{2} (logx)}{2} + \frac{ {x}^{2} }{4} - c}$

$\bold{⟹\frac{ {x}^{2} }{2} {(logx)}^{2} - \frac{ {x}^{2} (logx)}{2} + \frac{ {x}^{2} }{4} + c}$

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