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Index stranica MATEMATIČKA ANALIZA INTEGRALI

Tri neodređena integrala

[inlmath]\int xe^x\mathrm dx[/inlmath]

Tri neodređena integrala

Postod slavonija035 » Nedelja, 09. Jun 2013, 11:23

Riješite neodređeni integral:
[dispmath]1.\;\int\frac{x^2-3}{\sqrt{x+1}}\mathrm dx[/dispmath][dispmath]2.\;\int\frac{\ln\left(2x+1\right)}{\sqrt{2x+1}}\mathrm dx[/dispmath][dispmath]3.\;\int\frac{2x+2}{\left(x+2\right)\left(x^2+2\right)}\mathrm dx[/dispmath]
može pomoć oko ova 3 zadatka?
 
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Re: Tri neodređena integrala

Postod Daniel » Nedelja, 09. Jun 2013, 17:24

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]
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
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Re: Tri neodređena integrala

Postod Daniel » Nedelja, 09. Jun 2013, 17:43

slavonija035 je napisao:[dispmath]2.\;\int\frac{\ln\left(2x+1\right)}{\sqrt{2x+1}}\mathrm dx[/dispmath]

[dispmath]\int\frac{\ln\left(2x+1\right)}{\sqrt{2x+1}}\mathrm dx=\int\frac{\ln\left(\sqrt{2x+1}\right)^2}{\sqrt{2x+1}}\mathrm dx=\int\frac{2\ln\sqrt{2x+1}}{\sqrt{2x+1}}\mathrm dx=[/dispmath]
Smena
[inlmath]\sqrt{2x+1}=t\\
2x+1=t^2\\
x=\frac{t^2-1}{2}\\
\mathrm dx=t\mathrm dt[/inlmath]
[dispmath]=2\int\frac{\ln t}{\cancel t}\cancel t\mathrm dt=2\int\ln t\;\mathrm dt=[/dispmath]
[inlmath]u=\ln t\\
\mathrm du=\frac{\mathrm dt}{t}\\
\mathrm dv=\mathrm dt\\
v=t[/inlmath]
[dispmath]=2t\ln t-2\int\cancel t\frac{\mathrm dt}{\cancel t}=2t\ln t-2t+c=2t\left(\ln t-1\right)+c=t\left(\ln t^2-2\right)+c=[/dispmath][dispmath]=\sqrt{2x+1}\left(\ln\left|2x+1\right|-2\right)+c[/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
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Re: Tri neodređena integrala

Postod Daniel » Nedelja, 09. Jun 2013, 19:07

slavonija035 je napisao:[dispmath]3.\;\int\frac{2x+2}{\left(x+2\right)\left(x^2+2\right)}\mathrm dx[/dispmath]

Rastavljamo na parcijalne razlomke...
[dispmath]=\int\left(\frac{A}{x+2}+\frac{Bx+C}{x^2+2}\right)\mathrm dx=\int\frac{A\left(x^2+2\right)+\left(Bx+C\right)\left(x+2\right)}{\left(x+2\right)\left(x^2+2\right)}\mathrm dx=[/dispmath][dispmath]=\int\frac{Ax^2+2A+Bx^2+2Bx+Cx+2C}{\left(x+2\right)\left(x^2+2\right)}\mathrm dx=\int\frac{A\left(x^2+2\right)+\left(Bx+C\right)\left(x+2\right)}{\left(x+2\right)\left(x^2+2\right)}\mathrm dx=[/dispmath][dispmath]=\int\frac{Ax^2+2A+Bx^2+2Bx+Cx+2C}{\left(x+2\right)\left(x^2+2\right)}\mathrm dx=\int\frac{\left(A+B\right)x^2+\left(2B+C\right)x+\left(2A+2C\right)}{\left(x+2\right)\left(x^2+2\right)}\mathrm dx=[/dispmath]
[inlmath]\left.\begin{array}{l}
A+B=0\\
2B+C=2\\
2A+2C=2
\end{array}\right\}\quad\Rightarrow\quad\begin{array}{l}
A=-\frac{1}{3}\\
B=\frac{1}{3}\\
C=\frac{4}{3}
\end{array}[/inlmath]
[dispmath]=\int\left(\frac{-\frac{1}{3}}{x+2}+\frac{\frac{1}{3}x+\frac{4}{3}}{x^2+2}\right)\mathrm dx=-\frac{1}{3}\int\frac{\mathrm dx}{x+2}+\frac{1}{3}\int\frac{x+4}{x^2+2}\mathrm dx=[/dispmath][dispmath]=-\frac{1}{3}\ln\left|x+2\right|+\frac{1}{3}\int\frac{x\mathrm dx}{x^2+2}+\frac{4}{3}\int\frac{\mathrm dx}{x^2+2}=[/dispmath][dispmath]=-\frac{1}{3}\ln\left|x+2\right|+\frac{1}{6}\int\frac{\mathrm d\left(x^2+2\right)}{x^2+2}+\frac{2}{3}\int\frac{\mathrm dx}{\left(\frac{x}{\sqrt 2}\right)^2+1}=[/dispmath][dispmath]=-\frac{1}{3}\ln\left|x+2\right|+\frac{1}{6}\ln\left(x^2+2\right)+\frac{2\sqrt 2}{3}\int\frac{\mathrm d\left(\frac{x}{\sqrt 2}\right)}{\left(\frac{x}{\sqrt 2}\right)^2+1}=[/dispmath][dispmath]=-\frac{1}{3}\ln\left|x+2\right|+\frac{1}{6}\ln\left(x^2+2\right)+\frac{2\sqrt 2}{3}\mathrm{arctg}\frac{x\sqrt 2}{2}[/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
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