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Question Number 201184 by Calculusboy last updated on 01/Dec/23

Answered by Sutrisno last updated on 01/Dec/23

$$misal:\sqrt{2x}+4=u\to dx=\sqrt{2x}du$$$$=\int \frac{\sqrt{2x}}{u}.\sqrt{2x}du$$$$=\int \frac{{(u-4)}^2}{u}du$$$$=\int \frac{u^2-8u+16}{u}du$$$$=\int u-8+\frac{16}{u}du$$$$=\frac{1}{2}{u}^2-8u+16lnu+c$$$$=\frac{1}{2}{(\sqrt{2x}+4)}^2-8(\sqrt{2x}+4)+16ln(\sqrt{2x}+4)+c$$

Commented by Calculusboy last updated on 01/Dec/23

$$\mathit{n}\mathit{i}\mathit{c}\mathit{e}\mathit{s}\mathit{o}\mathit{l}\mathit{u}\mathit{t}\mathit{i}\mathit{o}\mathit{n}\mathit{s}\mathit{i}\mathit{r}$$

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