Question

(dy/dx)^3 - d^2y/dx^2 - dy/(dx) =0​

Answer 1

Step-by-step explanation:

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

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

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

$\huge\tt\underline\blue{❯Answer❮</p><p> }$

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

Let 1+logx=t

Differentiating both sides w.r.t.x

$⟹</p><p>0 + \frac{1}{x} = \frac{dt}{dx}$

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

$dx = xdt$

Integrating function:-

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

Putting 1+logx & dx =xdt

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

$= log |t| + c$

Put t=1+logx

$= log |1 + logx| + c$

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Answer 2

$\huge\tt\underline\purple{Answer</p><p> }$

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

Let 1+logx=t

Differentiating both sides w.r.t.x

$⟹</p><p>0 + \frac{1}{x} = \frac{dt}{dx}$

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

$dx = xdt$

Integrating function:-

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

Putting 1+logx & dx =xdt

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

$= log |t| + c$

Put t=1+logx

$= log |1 + logx| + c$