Question

$\int\ln(x+ \sqrt{x^2+1} ) \, dx$
Am rezolvat pana am ajuns la $x*ln(x+ \sqrt{x^2+1} )-arctgx*x- \int {arctgx} \, dx$ , ce mai pot face de aici ?

Answer 1

$\int\limits {\ln (x+\sqrt{x^2+1})} \, dx = \int\limits {x'\cdot \ln (x+\sqrt{x^2+1})} \, dx= \\ \\ = x\cdot \ln (x+\sqrt{x^2+1})-\int\limits {x\cdot \big(\ln (x+\sqrt{x^2+1})\big)'} \, dx\overset{(*)}{=}\\ \\ \Big(\quad\int\limits {\dfrac{1}{\sqrt{x^2+1}}} \, dx=\ln (x+\sqrt{x^2+1})+C\Big \rightarrow este chiar formula \quad \Big) \\ \\ \Leftrightarrow \Big(\quad\big(\ln (x+\sqrt{x^2+1})\big)' = \dfrac{1}{\sqrt{x^2+1}}\quad \Big)$

$\overset{(*)}{=} x\cdot \ln (x+\sqrt{x^2+1})-\int\limits {x\cdot\dfrac{1}{\sqrt{x^2+1}} \, dx= \\ \\ = x\cdot \ln (x+\sqrt{x^2+1}) - \int\limits {\dfrac{x}{\sqrt{x^2+1}} \, dx= \\ \\ \\ = x\cdot \ln (x+\sqrt{x^2+1}) - \int\limits {\big(\sqrt{x^2+1}\big)'} \, dx= \\ \\ = x\cdot \ln (x+\sqrt{x^2+1}) -\sqrt{x^2+1}+C$

$\Rightarrow \boxed{\int\limits {\ln (x+\sqrt{x^2+1})} \, dx = x\cdot \ln (x+\sqrt{x^2+1}) -\sqrt{x^2+1}+C}$