matematikk.net

[symbol:integral] tan^2 u du
og

[symbol:integral] ((z^(5/3)*(3 [symbol:rot] z))/(6 [symbol:rot] z^7) dz
og
[symbol:integral] dx/3 [symbol:rot] (5-7x)

[tex]1) \;\; \int \tan^2 u \, du \;=\; \int \frac{\sin^2u}{\cos^2u} \, du \;=\;\int \frac{1 \:-\: \cos^2u}{\cos^2u} \, du \;=\;\int \frac{1}{\cos^2u} \:-\: 1\, du \;=\; \tan u \:-\: u \:+\: C_1.[/tex]

[tex]2) \;\; \int \frac{z^{5/3} \, \cdot \, 3\sqrt{z}}{6\sqrt{z^7}} \, dz \;=\; \int \frac{z^{5/3}\, \cdot \, z^{1/2}}{2z^{7/2}} \, dz \;=\; \int \frac{z^{5/3 \:+\: 1/2 \:-\: 7/2}}{2} \, dz \;=\; \int \frac{z^{-4/3}}{2} \, dz \;=\; - \, \frac{3}{2} \, z^{-1/3} \:+\: C_2.[/tex]

[tex]3) \;\; \int \frac{dx}{\sqrt{5 \:-\: 7x}} \;=\; \int \frac{1}{\sqrt{u}} \: \frac{du}{-7} \;=\; \int \frac{u^{-1/2}}{-7} \, du \;=\; - \, \frac{2}{7} \, u^{1/2} \:+\: C_3 \;=\; - \, \frac{2}{7} \, \sqrt{5 \:-\: 7x} \:+\: C_3.[/tex]
[tex]\;\;\;\;\;[/tex]Anvender substitusjonen u = 5 - 7x.

NB: Her er C[sub]1[/sub], C[sub]2[/sub] og C[sub]3[/sub] vilkårlige konstanter.

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