Dva integrala racionalnih funkcija

PostPoslato: Nedelja, 09. Jun 2013, 11:57
od slavonija035
[dispmath]\begin{array}{lr}
7.\;\int\frac{2x-3}{5x-4}\mathrm dx & \left[\frac{2}{5}x-\frac{7}{25}\ln\left|x-\frac{4}{5}\right|+c\right] \\
10.\;\int\frac{5x-7}{x\left(2x^2-4x-6\right)}\mathrm dx\qquad & \left[\frac{7}{6}\ln\left|x\right|-\frac{3}{2}\ln\left|x+1\right|+\frac{1}{3}\ln\left|x-3\right|+c\right]
\end{array}[/dispmath]
prvi zadatak sam pokušao riješiti tako da sam izlučio [inlmath]2[/inlmath] iz brojnika i [inlmath]5[/inlmath] iz nazivnika ali ne mogu dobiti rješenje...
a ovaj drugi nemam ideju nažalost

Re: Dva integrala racionalnih funkcija

PostPoslato: Nedelja, 09. Jun 2013, 14:40
od Daniel
slavonija035 je napisao:[dispmath]7.\;\int\frac{2x-3}{5x-4}\mathrm dx[/dispmath]

Pa, dobro si krenuo. :)
[dispmath]=\frac{2}{5}\int\frac{x-\frac{3}{2}}{x-\frac{4}{5}}\mathrm dx=\frac{2}{5}\int\frac{10x-15}{10x-8}\mathrm dx=\frac{2}{5}\int\frac{10x-8-7}{10x-8}\mathrm dx=\frac{2}{5}\int\left(1-\frac{7}{10x-8}\right)\mathrm dx=[/dispmath][dispmath]=\frac{2}{5}\int\mathrm dx-\frac{14}{5}\int\frac{\mathrm dx}{10x-8}=\frac{2}{5}x-\frac{14}{50}\int\frac{\mathrm d\left(10x-8\right)}{10x-8}=\frac{2}{5}x-\frac{7}{25}\ln\left|10x-8\right|+c=[/dispmath][dispmath]=\frac{2}{5}x-\frac{7}{25}\ln\left|10\left(x-\frac{8}{10}\right)\right|+c=\frac{2}{5}x-\frac{7}{25}\left(\ln 10+\ln\left|x-\frac{4}{5}\right|\right)+c=[/dispmath][dispmath]=\frac{2}{5}x-\frac{7}{25}\ln 10-\frac{7}{25}\ln\left|x-\frac{4}{5}\right|+c=\frac{2}{5}x-\frac{7}{25}\ln\left|x-\frac{4}{5}\right|-\cancelto{c}{\enclose{circle}{\frac{7}{25}\ln 10+c}}=[/dispmath][dispmath]=\frac{2}{5}x-\frac{7}{25}\ln\left|x-\frac{4}{5}\right|+c[/dispmath]

slavonija035 je napisao:[dispmath]10.\;\int\frac{5x-7}{x\left(2x^2-4x-6\right)}\mathrm dx[/dispmath]

E ovde treba da primeniš metodu parcijalnih razlomaka. Prvo rastavimo trinom [inlmath]\left(2x^2-4x-6\right)[/inlmath]:
[inlmath]2x^2-4x-6=2\left(x^2-2x-3\right)\\
x_{1,2}=\frac{2\pm\sqrt{4+12}}{2}=\frac{2\pm 4}{2}=1\pm 2\\
x_1=-1\quad x_2=3\\
\Rightarrow\quad 2x^2-4x-6=2\left(x+1\right)\left(x-3\right)[/inlmath]
[dispmath]=\frac{1}{2}\int\frac{5x-7}{x\left(x+1\right)\left(x-3\right)}\mathrm dx=\frac{1}{2}\int\left(\frac{A}{x}+\frac{B}{x+1}+\frac{C}{x-3}\right)\mathrm dx=[/dispmath]
[dispmath]=\frac{1}{2}\int\frac{A\left(x+1\right)\left(x-3\right)+Bx\left(x-3\right)+Cx\left(x+1\right)}{x\left(x+1\right)\left(x-3\right)}\mathrm dx=[/dispmath][dispmath]=\frac{1}{2}\int\frac{Ax^2-2Ax-3A+Bx^2-3Bx+Cx^2+Cx}{x\left(x+1\right)\left(x-3\right)}\mathrm dx=[/dispmath][dispmath]=\frac{1}{2}\int\frac{\left(A+B+C\right)x^2+\left(-2A-3B+C\right)x-3A}{x\left(x+1\right)\left(x-3\right)}\mathrm dx=[/dispmath]
[inlmath]\left.\begin{array}{l}
A+B+C=0\\
-2A-3B+C=5\\
-3A=-7
\end{array}\right\}\quad\Rightarrow\quad\begin{array}{l}
A=\frac{7}{3}\\
B=-3\\
C=\frac{2}{3}
\end{array}[/inlmath][dispmath]=\frac{1}{2}\int\left(\frac{\frac{7}{3}}{x}+\frac{-3}{x+1}+\frac{\frac{2}{3}}{x-3}\right)\mathrm dx=\frac{7}{6}\int\frac{\mathrm dx}{x}-\frac{3}{2}\frac{\mathrm dx}{x+1}+\frac{1}{3}\frac{\mathrm dx}{x-3}=[/dispmath][dispmath]=\frac{7}{6}\ln\left|x\right|-\frac{3}{2}\ln\left|x+1\right|+\frac{1}{3}\ln\left|x-3\right|+c[/dispmath]