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Solve the following integrals by partial fractions or various changes of variables, if necessary. \(\int \frac {1-\sqrt{3x+2}}{1+\sqrt{3x+2}}dx\) |
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Answer» I = \(\int \frac {1-\sqrt{3x+2}}{1+\sqrt{3x+2}}dx\) Let 3x + 2 = t2 ∴ 3dx = 2+dt dx = \(\frac {2t}{3}dt\) ∴ I = \(\int \frac {1-t}{1+t}. \frac {2t}{3}dt\) = \(\frac 23 \int \frac {t-t^2}{1+t}dt\) = \(\frac 23 \int {t-t^2} - \frac {2}{1+t}dt\) = \(\frac 23[\frac {-t^2}{2} + 2t - 2log[1+t]]+c\) = \(\frac 13[t^2+4t-4log[1+t]] + c\) = \(\frac 13 [-(3x+2) + 4\sqrt{3x+2} - 4log [1t\sqrt{3x+2}]]+c\) (by putting t = \(\sqrt{3x+2}\)) ∴ \(\int \frac {1-\sqrt{3x+2}}{1+\sqrt{3x+2}}dx \) \(= \frac 13 [-(3x+2) +4\sqrt{3x+2}-4log [1+\sqrt3x+2]]+c\) |
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