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Question Number 208540 by otchereabdullai@gmail.com last updated on 17/Jun/24

Answered by Sutrisno last updated on 18/Jun/24

$$misal:3x=2tan\theta$$$$x=\frac{2}{3}tan\theta \to dx=\frac{2}{3}{sec}^2\theta d\theta$$$$=\int \frac{\frac{2}{3}tan\theta -3}{\sqrt{{\left(2tan\theta \right)}^2+4}}.\frac{2}{3}{sec}^2\theta d\theta$$$$=\int \frac{\frac{2}{3}tan\theta -3}{\sqrt{4({tan}^2\theta +1)}}.\frac{2}{3}{sec}^2\theta d\theta$$$$=\int \frac{\frac{2}{3}tan\theta -3}{2sec\theta }.\frac{2}{3}{sec}^2\theta d\theta$$$$=\int \frac{1}{3}(\frac{2}{3}tan\theta -3)sec\theta d\theta$$$$=\int \frac{1}{3}(\frac{2}{3}tan\theta sec\theta -3sec\theta )d\theta$$$$=\frac{1}{3}(\frac{2}{3}sec\theta -3ln(sec\theta +tan\theta ))+c$$$$=\frac{1}{3}(\frac{2}{3}\frac{\sqrt{9x^2+4}}{2}-3ln(\frac{\sqrt{9x^2+4}}{2}+\frac{3x}{2}))+c$$

Answered by Frix last updated on 18/Jun/24

$$\int \frac{x-3}{\sqrt{9x^2+4}}dx\stackrel{t=\frac{3x+\sqrt{9x^3+4}}{2}}{=}\int (\frac{1}{9}-\frac{1}{t}-\frac{1}{9t^2})dt=$$$$=\frac{t}{9}-\ln t+\frac{1}{9t}=\frac{t^2+1}{9t}-\ln t=$$$$=\frac{\sqrt{9x^2+4}}{9}-\ln (3x+\sqrt{9x^2+4})+C$$

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