calculate-0-1-ln-1-x-x-1-2-dx-

Question Number 91271 by mathmax by abdo last updated on 29/Apr/20

c a l c u l a t e \int_{0}^{1} \frac{l n (1 + x)}{{(x + 1)}^{2}} d x

Commented by mathmax by abdo last updated on 29/Apr/20

I = \int_{0}^{1} \frac{l n (1 + x)}{{(x + 1)}^{2}} d x b y p s r t s w e g e t

I = {[- \frac{1}{x + 1} l n (x + 1)]}_{0}^{1} + \int_{0}^{1} \frac{l n (x + 1)}{x + 1} d x

= - \frac{1}{2} l n (2) + \int_{1}^{2} \frac{l n t}{t} d t (x + 1 = t)

\int_{1}^{2} \frac{l n (t)}{t} d t =_{l n t = u} \int_{0}^{l n (2)} \frac{u}{e^{u}} e^{u} d u = \int_{0}^{l n (2)} u d u = {[\frac{u^{2}}{2}]}_{0}^{l n (2)}

= \frac{1}{2} {l n}^{2} (2) \Rightarrow I = - \frac{1}{2} l n (2) + \frac{1}{2} {l n}^{2} (2) .

Answered by M±th+et+s last updated on 29/Apr/20

I = \int_{0}^{1} \frac{l n (1 + x)}{x^{2} + 1} d x

p u t x = t a n (u) d x = {s e c}^{2} (u)

= \int_{0}^{\frac{π}{4}} \frac{l n (t a n (u) + 1)}{{t a n}^{2} u + 1} {s e c}^{2} (u) d u

= \int_{0}^{\frac{π}{4}} l n (t a n (u) + 1) d u

= \int_{0}^{\frac{π}{4}} l n (\frac{s i n u + c o s u}{c o s u}) d u

\int_{0}^{\frac{π}{4}} l n (s i n u + c o s u) - l n (c o s u) d u

\int_{0}^{\frac{π}{4}} l n (\sqrt{2} (c o s (\frac{π}{4} - u)) - l n (c o s u) d u

\int_{0}^{\frac{π}{4}} l n (\sqrt{2}) + l n (c o s (\frac{π}{4} - u)) - l n (c o s u) d u

J = \int_{0}^{\frac{π}{4}} l n (c o s (\frac{π}{4} - u)) d u

\frac{π}{4} - u = t d u = - d t

t h e n J = \int_{\frac{π}{4}}^{0} - l n (c o s (t)) d t \int_{0}^{\frac{π}{4}} l n (c o s (t)) d t

\int_{0}^{\frac{π}{4}} l n (c o s (t)) d t = \int_{0}^{\frac{π}{4}} l n (c o s (u)) d u

t h e n \int_{0}^{\frac{π}{4}} l n (\sqrt{2}) + l n (c o s (u)) - l n (c o s (u)) d u

= \int_{0}^{\frac{π}{4}} l n (\sqrt{2}) d u = l n \sqrt{2} \frac{π}{4} = \frac{π}{8} l n (2)

Commented by mathmax by abdo last updated on 29/Apr/20

t h a n k y o u s i r .

Commented by M±th+et+s last updated on 29/Apr/20

y o u a r e w e l c o m e

Commented by mathmax by abdo last updated on 29/Apr/20

l o o k s i r t h e i n t e g r a l i s \int_{0}^{1} \frac{l n (1 + x)}{{(1 + x)}^{2}} d x n o t \int_{0}^{1} \frac{l n (1 + x)}{1 + x^{2}} d x

a n y w a y t h a n k f o r y o u r w o r k \dots

Commented by M±th+et+s last updated on 29/Apr/20

s o r r y, i {d i d n}^{'} t n o t i c e