Question

integrate the function :
$\frac{1}{x + xlogx}$

Answer 1

Answer:

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

$\red{\bold{\underline{\underline{❥Question᎓}}}}$integrate the function :

$\frac{1}{x + xlogx}$

$\huge\huge\tt\blue{「Answer」</p><p> }$

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$⟹ \frac{1}{x + xlogx} = \frac{1}{x(1 + logx)} </p><p>$

$\bold{\red{Let 1+logx=t}}$

$\bold{Differentiating\: both \:sides \:w.r.t.x}[/t ex]</p><p>[tex]⟹</p><p>0 + \frac{1}{x} = \frac{dt}{dx}$

$⟹</p><p> \frac{1}{x} = \frac{dt}{dx}$

$dx = xdt$

$\bold{Integrating function:-}$

$⟹∫ \frac{1}{x + xlogx} dx = ∫ \frac{1}{x(1 + logx)} dx</p><p>$

$\bold{Putting \:1+logx\: &\: dx =xdt}$

$= ∫ \frac{1}{x(t)} dt \times x = ∫ \frac{1}{t} dt$

$= log |t| + c$

$\bold{Put\: t=1+logx}$

$= log |1 + logx| + c$tt

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Answer 2

According to the question

$⟹ \frac{1}{x + xlogx} = \frac{1}{x(1 + logx)} \\ \\ </p><p></p><p>\bold{\red{Let 1+logx=t}} \\ \\ </p><p></p><p>\bold{Differentiating\: both \:sides \:w.r.t.x} \\ \\ [/t ex] [tex]⟹ 0 + \frac{1}{x} = \frac{dt}{dx}$ $</p><p></p><p>⟹ \frac{1}{x} = \frac{dt}{dx} \\ \\ </p><p></p><p></p><p></p><p>\bold{Integrating function:-} \\ \\ </p><p></p><p>⟹∫ \frac{1}{x + xlogx} dx = ∫ \frac{1}{x(1 + logx)} \\ \\ </p><p></p><p>\bold{Putting \:1+logx\: &\: dx =xdt} \\ \\ </p><p></p><p>= ∫ \frac{1}{x(t)} dt \times x = ∫ \frac{1}{t} \\ \\ </p><p></p><p>= log |t| + c=log∣t∣+c \\ \\ </p><p></p><p>\bold{Put\: t=1+logx} \\ \\ </p><p></p><p>= log |1 + logx| + c=log∣1+logx∣+c tt</p><p></p><p>$