Math, asked by Anonymous, 7 months ago

Integrate by parts:-
x {(logx)}^{2}
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Answered by Anonymous
73

\huge{\bold☘}\mathfrak\pink{\bold{\underline{{ ℘ɧεŋσɱεŋศɭ}}}}{\bold☘}

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

x {(logx)}^{2}

\huge\tt{\boxed{\overbrace{\underbrace{\blue{Answer</p><p> }}}}}

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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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