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:

f(x) = kx³ – 8x² + 5

Roots are α – β , α & α +β

Sum of roots = – (-8)/k

Sum of roots = α – β + α + α +β = 3α

= 3α = 8/k

= k = 8/3α

or we can solve as below

f(x) = (x – (α – β)(x – α)(x – (α +β))

= (x – α)(x² – x(α+β + α – β) + (α² – β²))

= (x – α)(x² – 2xα + (α² – β²))

= x³ – 2x²α + x(α² – β²) – αx² +2α²x – α³ + αβ²

= x³ – 3αx² + x(3α² – β²) + αβ² – α³

= kx³ – 3αkx² + xk(3α² – β²) + k(αβ² – α³)

comparing with

kx³ – 8x² + 5

k(3α² – β²) = 0 => 3α² = β²

k(αβ² – α³) = 5

=k(3α³ – α³) = 5

= k2α³ = 5

3αk = 8 => k = 8/3α

(8/3α)2α³ = 5

=> α² = 15/16

=> α = √15 / 4