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Integration! \[\int \frac{x+3}{\sqrt{x+6}}dx\]
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So lemme see. \(u = x + 6\) \(du = dx\) And that means \(x + 3 = u - 3\) \[\int \frac{u-3}{\sqrt{u}}\] So far so good?
ok
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\[\int \frac{u}{\sqrt{u}} - \frac{3}{\sqrt{u}}\]\[\int u^{1/2} - 3u^{-1/2}\]\[\frac{2}{3}u^{3/2} - 6u^{1/2} + C\]
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I guess it's just\[\frac{2}{3}(x + 6)^{3/2} - 6(x + 6)^{1/2} + C\]Can I clean that up any?
2(x+6)^1/2 (1/3 (x+6)^3-3) +C how ugly it is , hahahaha
It's a good idea though. Maybe factor something out. \[\sqrt{x + 6}\left(\frac{2}{3}(x + 6) - 6\right) + C\]\[\sqrt{x + 6}\left(\frac{2x}{3} - 2\right) + C\]\[\frac{2}{3}\sqrt{x + 6}(x - 3) + C\] Well that looks nicer.
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