0-1-x-2-1-lnx-dx-

Question Number 47770 by tanmay.chaudhury50@gmail.com last updated on 14/Nov/18

\int_{0}^{1} \frac{x^{2} - 1}{l n x} d x

Commented by maxmathsup by imad last updated on 14/Nov/18

y o u a r e w e l c o m e s i r .

Commented by maxmathsup by imad last updated on 14/Nov/18

l e t I = \int_{0}^{1} \frac{x^{2} - 1}{l n (x)} d x c h a n g e m e n t l n (x) = - t g i v e x = \overset{- t}{e}

I = \int_{0}^{+ \infty} \frac{e^{- 2 t} - 1}{- t} e^{- t} d t = \int_{0}^{\infty} \frac{e^{- t} - e^{- 3 t}}{t} d t l e t d e t e r m i n e

f (x) = \int_{0}^{\infty} \frac{e^{- t} - e^{- 3 t}}{t} e^{- x t} d t w i t h x ⩾ 0 w e h s v e

f^{^{'}} (x) = - \int_{0}^{\infty} (e^{- t} - e^{- 3 t}) e^{- x t} d t = - \int_{0}^{\infty} (e^{- (x + 1) t} - e^{- (x + 3) t}) d t

= - {[- \frac{1}{x + 1} e^{- (x + 1) t} + \frac{1}{x + 3} e^{- (x + 3) t}]}_{t = 0}^{+ \infty} = - (\frac{1}{x + 1} - \frac{1}{x + 3}) \Rightarrow f (x) = - l n (\frac{x + 1}{x + 3}) + λ

b u t \exists m > 0 / ∣ f (x) ∣⩽ m \int_{0}^{\infty} e^{- x t} d t = \frac{m}{x} \to 0 (x \to + \infty) \Rightarrow

λ = {l i m}_{x \to + \infty} (f (x) + l n (\frac{x + 1}{x + 3})) = 0 \Rightarrow f (x) = - l n (\frac{x + 1}{x + 3}) \Rightarrow f (x) = l n (\frac{x + 3}{x + 1})

I = f (0) = l n (3) \Rightarrow ★ I = l n (3) ★ .

Commented by tanmay.chaudhury50@gmail.com last updated on 14/Nov/18

t h a n k y o u s i r \dots

Answered by tanmay.chaudhury50@gmail.com last updated on 14/Nov/18

s o m e t r i c k s t o s o l v e \dots

I = \int_{0}^{1} \frac{x^{a} - 1}{l n x} d x

\frac{d I}{d a} = \int_{0}^{1} \frac{\partial (x^{a} - 1)}{\partial a} \times \frac{1}{l n x} d x

= \int_{0}^{1} \frac{x^{a} \times l n x}{l n x} d x

= \int_{0}^{1} x^{a} d x

=∣ \frac{x^{a + 1}}{a + 1} ∣_{0}^{1} = \frac{1}{a + 1}

\frac{d I}{d a} = \frac{1}{a + 1}

d I = \frac{d a}{a + 1}

I = l n (a + 1) + C

p u t a = 0 s o \int \frac{x^{a} - 1}{l n x} d x = 0

h e n c e I = l n (a + 1) + C

0 = l n (0 + 1) + C

C = 0

s o I = l n (a + 1)

s o a n s w e r f o r \int_{0}^{1} \frac{x^{2} - 1}{l n x} d x = l n (2 + 1) = l n 3