Evo, vjerovatno si vec rijesio
Ali ako nekome nekada bude trebalo
[dispmath]\int\limits_0^1\frac{x-1}{\sqrt{x}}\,\mathrm dx[/dispmath][dispmath]\int\frac{x-1}{\sqrt{x}}\,\mathrm dx[/dispmath][dispmath]\int\frac{x-1}{x^\frac{1}{2}}\,\mathrm dx[/dispmath][dispmath]\int\left(\frac{x}{x^\frac{1}{2}}-\frac{1}{x^\frac{1}{2}}\right)\mathrm dx[/dispmath][dispmath]\int x^\frac{1}{2}\,\mathrm dx-\int\frac{1}{x^\frac{1}{2}}\,\mathrm dx[/dispmath][dispmath]\left.\left(\frac{2x\sqrt{x}}{3}-2\sqrt{x}\right)\right|_0^1[/dispmath][dispmath]\left(\frac{2\cdot1\cdot\sqrt1}{3}-2\cdot\sqrt1\right)-\left(\frac{2\cdot0\cdot\sqrt0}{3}-2\cdot\sqrt0\right)[/dispmath][dispmath]\frac{2}{3}-2=-\frac{4}{3}[/dispmath] To je prvi
, a evo i drugog
[dispmath]\int\limits_0^1\frac{1}{x+5}\,\mathrm dx[/dispmath][dispmath]\int\frac{1}{x+5}\,\mathrm dx[/dispmath][dispmath]t=x+5[/dispmath][dispmath]\int\frac{1}{t}\,\mathrm dt[/dispmath][dispmath]\ln(|t|)[/dispmath][dispmath]\ln(|x+5|)\,\Big|_0^1[/dispmath][dispmath]\ln(1+5)-\ln(0+5)[/dispmath][dispmath]\ln(6)-\ln(5)[/dispmath][dispmath]\ln\left(\frac{6}{5}\right)[/dispmath] To je to, nadam se da nisam negdje pogrijesio