Question

anuj got 58 marks out of 75 in physics and 97 marks out of 120 in maths. In which of the two subjects his performance was better?​

Answer 1

Step-by-step explanation:

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

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

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

${\huge{\fcolorbox{aqua}{navy}{\fcolorbox{yellow}{blue}{\bf{\color{yellow}{Answer}}}}}}$

╔════════════════════════╗

_ _ _ _ _ _ _ _ _ _ _ _ _ _ _✍️

Given:-

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

Here ,this identity is used :-

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

╚════════════════════════╝

нσρє ıт нєłρs yσυ

_____________________

тнαηkyσυ