slavonija035 je napisao:[dispmath]1.\;\int\frac{x^2-3}{\sqrt{x+1}}\mathrm dx[/dispmath]
[dispmath]\int\frac{x^2-3}{\sqrt{x+1}}\mathrm dx=\int\frac{x^2+x-x-1-2}{\sqrt{x+1}}\mathrm dx=\int\frac{x\left(x+1\right)-\left(x+1\right)-2}{\sqrt{x+1}}\mathrm dx=[/dispmath][dispmath]=\int\left(x\sqrt{x+1}-\sqrt{x+1}-\frac{2}{\sqrt{x+1}}\right)\mathrm dx=\int x\sqrt{x+1}\mathrm dx-\int\sqrt{x+1}\;\mathrm dx-2\int\frac{\mathrm dx}{\sqrt{x+1}}=[/dispmath][dispmath]=\int x\sqrt{x+1}\mathrm dx-\frac{\left(x+1\right)^\frac{3}{2}}{\frac{3}{2}}-2\frac{\left(x+1\right)^\frac{1}{2}}{\frac{1}{2}}=\int x\sqrt{x+1}\mathrm dx-\frac{2}{3}\left(x+1\right)^\frac{3}{2}-4\sqrt{x+1}=[/dispmath]
Razlomak [inlmath]\int x\sqrt{x+1}\;\mathrm dx[/inlmath] rešavamo parcijalnom integracijom:
[inlmath]u=x\\
\mathrm du=\mathrm dx\\
\mathrm dv=\sqrt{x+1}\;\mathrm dx\\
v=\frac{2}{3}\left(x+1\right)^\frac{3}{2}[/inlmath]
[dispmath]=\frac{2}{3}x\left(x+1\right)^\frac{3}{2}-\frac{2}{3}\int\left(x+1\right)^\frac{3}{2}\;\mathrm dx-\frac{2}{3}\left(x+1\right)^\frac{3}{2}-4\sqrt{x+1}=[/dispmath][dispmath]=\frac{2}{3}x\left(x+1\right)^\frac{3}{2}-\frac{2}{3}\cdot\frac{2}{5}\left(x+1\right)^\frac{5}{2}-\frac{2}{3}\left(x+1\right)^\frac{3}{2}-4\sqrt{x+1}+c=[/dispmath][dispmath]=\frac{\sqrt{x+1}}{15}\left[10x\left(x+1\right)-4\left(x+1\right)^2-10\left(x+1\right)-60\right]+c=[/dispmath][dispmath]=\frac{2\sqrt{x+1}}{15}\left[5x\left(x+1\right)-2\left(x+1\right)^2-5\left(x+1\right)-30\right]+c=[/dispmath][dispmath]=\frac{2\sqrt{x+1}}{15}\left(3x^2-4x-37\right)+c[/dispmath]