całka
: 24 mar 2009, o 11:59
\(\displaystyle{ \int ln(x + \sqrt{1 + x^2} )dx}\)
przez czesci
\(\displaystyle{ f=\ln (x+ \sqrt{1+x^{2}}) \ \ , \ \ \frac{df}{dx}=\frac{1}{\sqrt{1+x^{2}}}}\)
\(\displaystyle{ \frac{dg}{dx}=1 \ \ \ , \ \ \ g=x}\)
\(\displaystyle{ \int \ln (x+ \sqrt{1+x^{2}})dx= x \cdot \ln (x+ \sqrt{1+x^{2}}) - \int \frac{x}{\sqrt{1+x^{2}}} = x* arcsinh (x) - \sqrt{1+x^{2}} +C}\)