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Jeg fant denne, veldig morsom oppgave.

[tex]\int \left(\frac{x}{\sqrt {x+1}}\right)\rm{d}x[/tex]

[tex]u = x + 1, \ \frac{du}{dx} = 1, \ du = dx[/tex]

[tex]I = \int \frac{x}{\sqrt{x + 1}}dx = \int \frac{u - 1}{\sqrt u} du = \int \left(\frac{u}{sqrt {u}} - \frac{1}{sqrt{u}}\right) du = \int \sqrt u du - \int \frac{1}{\sqrt u} du = I_1 - I_2[/tex]

Så er vel resten bare potenskjøring

[tex]I_1 = \int \sqrt u du = \int u^{\frac{1}{2}} = \frac{1}{\frac{1}{2} + 1}u^{\frac{1}{2} + 1} = \frac{2}{3}u^{\frac{3}{2}} = \frac{2}{3}\sqrt{u^3} + C_1[/tex]

[tex]I_2 = \int \frac{1}{\sqrt{u}} du = \int u^{-\frac{1}{2}} du = \frac{1}{-\frac{1}{2} + 1} u^{-\frac{1}{2} + 1} = 2u^{\frac{1}{2}} = 2\sqrt u + C_2[/tex]

[tex]I = I_1 - I_2 = \frac{2}{3}\sqrt{u^3} + C_1 - (2\sqrt u + C_2) = \frac{2}{3}\sqrt{(x + 1)^3} - 2\sqrt{x+1} + C, \ C = C_1 - C_2[/tex]

Ser bra ut det Vektormannen.

Evt bruk substitusjonen u[sup]2[/sup] = x+1

Her kommer en artig oppfølger.

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

[tex]I=\int \frac1x (ln\,x)^2\rm{d}x \\ u=\ln\,x \\ \rm{d}u=\frac{\rm{d}x}{x} \\ I=\int u^2\rm{d}u=\frac13u^3+C=\underline{\underline{\frac13\ln ^3 x+C}}[/tex]

espen180 skrev:[tex]I=\int \frac1x (ln\,x)^2\rm{d}x \\ u=\ln\,x \\ \rm{d}u=\frac{\rm{d}x}{x} \\ I=\int u^2\rm{d}u=\frac13u^3+C=\underline{\underline{\frac13\ln ^3 x+C}}[/tex]

Flott!
Nå kan du ta det via delvis integrasjon også :]

[tex]I=\int\frac{\ln\,x}{x}\ln\,x\rm{d}x \\ v=\ln\,x \,\,\, v^\prime=\frac1x \\ u=\frac12\ln^2x \,\,\, u^\prime=\frac{\ln\,x}{x} \\ I=\frac12\ln^3x-\frac12\int \frac{\ln^2x}{x}\rm{d}x \\ \frac32I=\frac12\ln^3x+C \\ \underline{\underline{I=\frac13\ln^3x+C}}[/tex]