Question

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

Answer 1

Step-by-step explanation:

Step-by-step explanation:

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

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

$\huge\tt\underline\pink{꧁Answer꧂</p><p> }$

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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

Step-by-step explanation:

Step-by-step explanation:

given, \int{\frac{1}{x(logx)^m}}\,dx∫

x(logx)

m

1

dx , x >0

let logx = t

differentiate both sides,

1/x dx = dt , put it above Integration.

$\sf \: , \int{\frac{1}{x(logx)^m}}\,dx=\int{\frac{dt}{t^m}}∫$

=

$\sf \: t^{-m}$

$\sf\frac{t^{1-m}}{1-m}$

now, put t = logx

so, I =

$\bf{\frac{(logx)^{1-m}}{1-m}}$