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Ovi korisnici su zahvalili autoru
Daniel za post:
Milovan
Reputacija: 4.55%
od Daniel » Petak, 27. Decembar 2013, 23:57
Evo ovako:
[dispmath]\int\frac{x}{\sqrt{x^2+x+2}}\mathrm dx=\frac{1}{2}\int\frac{2x+1-1}{\sqrt{x^2+x+2}}\mathrm dx=\frac{1}{2}\int\frac{2x+1}{\sqrt{x^2+x+2}}\mathrm dx-\frac{1}{2}\int\frac{1}{\sqrt{x^2+x+2}}\mathrm dx=[/dispmath][dispmath]=\frac{1}{2}\int\frac{\mathrm d\left(x^2+x+2\right)}{\sqrt{x^2+x+2}}-\frac{1}{2}\int\frac{1}{\sqrt{x^2+x+2}}\mathrm dx=[/dispmath][dispmath]=\frac{1}{2}\int\left(x^2+x+2\right)^{-\frac{1}{2}}\mathrm d\left(x^2+x+2\right)-\frac{1}{2}\int\frac{1}{\sqrt{x^2+x+2}}\mathrm dx=[/dispmath][dispmath]=\cancel{\frac{1}{2}}\frac{\left(x^2+x+2\right)^\frac{1}{2}}{\cancel{\frac{1}{2}}}-\frac{1}{2}\int\frac{1}{\sqrt{x^2+x+2}}\mathrm dx=\sqrt{x^2+x+2}-\frac{1}{2}\int\frac{1}{\sqrt{x^2+x+2}}\mathrm dx[/dispmath]
I do not fear death. I had been dead for billions and billions of years before I was born, and had not suffered the slightest inconvenience from it. – Mark Twain