Stemmer det der, du har jo kommet frem til riktig resultat
Jeg gjorde det sånn:
[tex]I = \int \sqrt{1 + \ln (x)} {\rm d}x[/tex]
Delvis
[tex]u^\prime = 1[/tex], [tex]v = \sqrt{1 + \ln (x)}[/tex]
[tex]u = x[/tex], [tex]v^\prime = \frac{1}{2x\sqrt{1 + \ln (x)}}[/tex]
[tex]I = x\sqrt{1 + \ln (x)} - \int \frac{{\rm d}x}{2\sqrt{1 + \ln (x)}}[/tex]
[tex]I = x\sqrt{1 + \ln (x)} - J[/tex], [tex]J = \int \frac{{\rm d}x}{2\sqrt{1 + \ln (x)}}[/tex]
Løser det siste integralet. Variabelskifte.
[tex]u = \sqrt{1 + \ln (x)[/tex], [tex]x = e^{u^2 - 1}[/tex]
[tex]u^\prime = \frac{1}{2x\sqrt{1 + \ln (x)}} = \frac{1}{2ue^{u^2 - 1}}[/tex]
[tex]{\rm d}x = 2ue^{u^2 - 1} {\rm d}u[/tex]
[tex]J = \int \frac{2ue^{u^2 - 1} {\rm d}u}{2u} = \int \frac{1}{e} e^{u^2} {\rm d}u = \frac{\sqrt{\pi}}{2e} \cdot \frac{2}{\sqrt{\pi}} \int e^{u^2} {\rm d}u[/tex]
[tex]J = \frac{\sqrt{\pi}}{2e} erfi(u) + C[/tex]
[tex]J = \frac{\sqrt{\pi}}{2e} erfi(\sqrt{1 + \ln (x)} ) + C[/tex]
[tex]I = x\sqrt{1 + \ln (x)} - \frac{\sqrt{\pi}}{2e} erfi(\sqrt{1 + \ln (x)} ) + C[/tex]