calculus-2-1-2-ln-x-2-x-1-x-dx-

Question Number 133369 by mnjuly1970 last updated on 21/Feb/21

c a l c u l u s (2) \dots . .

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

Commented by Dwaipayan Shikari last updated on 21/Feb/21

\frac{{l o g}^{2} (2)}{2}

Commented by mnjuly1970 last updated on 21/Feb/21

c o r r e c t \dots

Answered by mathmax by abdo last updated on 21/Feb/21

Φ = \int_{1}^{2} \frac{1}{x} \ln (\frac{x + 2}{x + 1}) dx we do the changement \frac{1}{x} = t \Rightarrow

Φ = \int_{1}^{\frac{1}{2}} tln (\frac{\frac{1}{t} + 2}{\frac{1}{t} + 1}) (- \frac{dt}{t^{2}}) = \int_{\frac{1}{2}}^{1} \frac{1}{t} \ln (\frac{1 + 2 t}{1 + t}) dt

= \int_{\frac{1}{2}}^{1} \frac{\ln (1 + 2 t)}{t} dt - \int_{\frac{1}{2}}^{1} \frac{\ln (1 + t)}{t} dt = H - K

K = {[lnt . \ln (1 + t)]}_{\frac{1}{2}}^{1} - \int_{\frac{1}{2}}^{1} \frac{\ln (t)}{t + 1} dt

= - \ln (\frac{1}{2}) \ln (\frac{3}{2}) - \int_{\frac{1}{2}}^{1} \frac{\ln (t)}{1 + t} dt and

\int_{\frac{1}{2}}^{1} \frac{\ln (t)}{1 + t} dt = \int_{\frac{1}{2}}^{1} \ln (t) \sum_{n = 0}^{\infty} {(- 1)}^{n} t^{n} dt

= \sum_{n = 0}^{\infty} {(- 1)}^{n} \int_{\frac{1}{2}}^{1} t^{n} \ln (t) dt = \sum_{n = 0}^{\infty} {(- 1)}^{n} u_{n}

u_{n} = {[\frac{t^{n + 1}}{n + 1} lnt]}_{\frac{1}{2}}^{1} - \int_{\frac{1}{2}}^{1} \frac{t^{n + 1}}{n + 1} \frac{dt}{t} = - \ln (\frac{1}{2}) \frac{1}{(n + 1) 2^{n + 1}} - \frac{1}{n + 1} \int_{\frac{1}{2}}^{1} t^{n} dt

= \frac{\ln (2)}{(n + 1) 2^{n}} - \frac{1}{{(n + 1)}^{2}} (1 - \frac{1}{2^{n + 1}}) \Rightarrow

\int_{\frac{1}{2}}^{1} \frac{lnt}{1 + t} dt = \ln 2 \sum_{n = 0}^{\infty} \frac{{(- 1)}^{n}}{(n + 1) 2^{n}} - \sum_{n = 0}^{\infty} \frac{{(- 1)}^{n}}{{(n + 1)}^{2}} + \sum_{n = 0}^{\infty} \frac{{(- 1)}^{n}}{{(n + 1)}^{2} . 2^{n + 1}}

rest calculus of those series \dots

H = \int_{\frac{1}{2}}^{1} \frac{\ln (1 + 2 t)}{t} dt =_{2 t = y} 2 \int_{1}^{2} \frac{\ln (1 + y)}{y} \frac{dy}{2}

=_{y = u + 1} \int_{o}^{1} \frac{\ln (u + 2)}{u + 1} du = \int_{0}^{1} \ln (2 + u) \sum_{n = 0}^{\infty} {(- 1)}^{n} u^{n} du

= \sum_{n = 0}^{\infty} {(- 1)}^{n} \int_{0}^{1} u^{n} \ln (2 + u) du = \sum_{n = 0}^{\infty} {(- 1)}^{n} v_{n}

v_{n} = {[\frac{u^{n + 1}}{n + 1} \ln (2 + u)]}_{0}^{1} - \int_{0}^{1} \frac{u^{n + 1}}{n + 1} \frac{du}{2 + u}

= \frac{\ln (3)}{n + 1} - \frac{1}{n + 1} \int_{0}^{1} \frac{u^{n + 1}}{u + 2} du = \dots . be continued \dots .

Answered by mnjuly1970 last updated on 22/Feb/21