matematikk.net

[tex]I=\int x^x\left(x(\ln x)^2+x\ln x\right)\rm{d}x[/tex]

[tex]I=\int x \ln x e^{x \ln x}(\ln x +1) \rm{d}x[/tex]

[tex]u=x \ln x \Rightarrow \frac{ \rm{d}u}{\rm{d} x } = \ln x +1[/tex]

[tex]I=\int u e^u \rm{d}u = ue^u-\int e^u \rm{d}u=e^u(u-1)+C=x^x(x \ln{x}-1)+C[/tex]

Oppfølger:

[tex]\int \frac{2x+1}{x^2-x+2} \rm{d}x[/tex]

sikkert ikke enkleste løsningen men

[tex]I=\int\frac{2x+1}{x^2-x+1}\quad\mathrm{d}x=\int\frac{2x+1}{\left(x-\frac{1}{2}\right)^2+\frac{7}{4}}\quad\mathrm{d}x[/tex]

[tex]u=\left(x-\frac{1}{2}\right)\Leftrightarrow x=u+\frac{1}{2}[/tex]
[tex]\mathrm{d}u=\mathrm{d}x[/tex]

[tex]I=\int\frac{2u+2}{u^2+\frac{7}{4}}\quad\mathrm{d}u[/tex]

[tex]I_1=\int\frac{2u}{u^2+\frac{7}{4}}\quad\mathrm{d}u[/tex]
[tex]I_2=\int\frac{2}{u^2+\frac{7}{4}}\quad\mathrm{d}u=\int\frac{2}{\frac{7}{4}\left(\frac{4}{7}u^2+1\right)}\quad\mathrm{d}u=\frac{8}{7}\int\frac{1}{\left(\frac{4}{7}u^2+1\right)}\quad\mathrm{d}u[/tex]

[tex]I_1:\quad v={u^2+\frac{7}{4}}\Rightarrow \mathrm{d}u=\frac{1}{2u}\mathrm{d}v\Rightarrow I_1=\int\frac{1}{v}\quad\mathrm{d}v=\ln(v)=\ln\left(u^2+\frac{7}{4}\right)=\ln\left(x^2-x+2\right)[/tex]

[tex]I_2:\quad z=\sqrt{\frac{4}{7}}u\Rightarrow \sqrt{\frac{7}{4}}\mathrm{d}z=\mathrm{d}u\Rightarrow I_2=\frac{8}{7}\sqrt{\frac{7}{4}}\int\frac{1}{z^2+1}\mathrm{d}z=\frac{4}{\sqrt{7}}\tan^{-1}(z)=\frac{4}{\sqrt{7}}\tan^{-1}\left(\sqrt{\frac{4}{7}}u\right)=\frac{4}{\sqrt{7}}\tan^{-1}\left(\frac{2}{\sqrt{7}}\left(x-\frac{1}{2}\right)\right)=\frac{4}{\sqrt{7}}\tan^{-1}\left(\frac{2x-1}{\sqrt{7}}\right)[/tex]

[tex]I=I_1+I_2=\ln\left(x^2-x+2\right)+\frac{4}{\sqrt{7}}\tan^{-1}\left(\frac{2x-1}{\sqrt{7}}\right)+C[/tex]

To nye oppføgere:

[tex]I_1=\int\sqrt{\frac{x^3}{1+x}}\rm{d}x[/tex]

og

[tex]I_2=\int\sqrt{\frac{x^3}{n+x}}\rm{d}x[/tex]

jeg orker rett og slett ikke skrive alt, men man kan begynne med denne substitusjonen:

[tex]x=\sinh^2(u)[/tex]
der
[tex]dx=2\sinh(u)\cosh(u)\,du[/tex]

og
[tex]x+1=\sinh^2(u)+1=\cosh^2(u)[/tex]

[tex]\large I_1=\int x\sqrt{\frac{x}{x+1}}\,dx=2\int \sinh^4(u)\,du[/tex]

dette må bearbeides videre...

svaret fås på Wolfram...

http://integrals.wolfram.com/index.jsp

Man kan bruteforce den med delvis:

[tex]\int \sinh^4 x\rm{d}x = \cosh x \sinh^3x-\int 3\sinh^2x \cosh^2x \rm{d}x=\cosh x \sinh^3x-\int 3\sinh^4x -3\sinh^2x \rm{d}x[/tex]

[tex]\int \sinh^4x \rm{d}x=\frac{1}{4}(\cosh x \sinh^3x+3\int \sinh^2x \rm{d}x)[/tex]

[tex]\int \sinh^2x \rm{d}x=\cosh x \sinh x - \int \cosh^2x \rm{d}x =\cosh x \sinh x - \int \sinh^2x+1 \rm{d}x[/tex]

[tex]\int \sinh^2x \rm{d}x=\frac{1}{2}(\cosh x \sinh x-x)[/tex]

[tex]\int \sinh^4x \rm{d}x=\frac{1}{8}(2\cosh x \sinh^3x+3\cosh x \sinh x-3x)[/tex]

Så er det bare å substituere tilbake.