Find the integral of: [g'(x)] / [g(x)]^2 I got as an anser: [g(x)] / 1/3[g(x)]^3 ...but I'm not sure if I broke some kind of rules.
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Originally Posted by WartonMorton Find the integral of: [g'(x)] / [g(x)]^2 I got as an anser: [g(x)] / 1/3[g(x)]^3 ...but I'm not sure if I broke some kind of rules. hi warton,.. i suggest you to use the substitution method $\displaystyle \int \frac{g'(x)}{g^2(x)}~dx = \int \frac{~d(g(x))}{g^2(x)} = - \frac{1}{g(x)} + C$
Originally Posted by WartonMorton Find the integral of: [g'(x)] / [g(x)]^2 I got as an anser: [g(x)] / 1/3[g(x)]^3 ...but I'm not sure if I broke some kind of rules. $\displaystyle u = g(x)$ therefore $\displaystyle du = g'(x) dx \: \: \rightarrow \: dx = \frac{du}{g'(x)}$ $\displaystyle \int \frac{g'(x)}{u^2\,g'(x)}\,du = \int \frac{du}{u^2}$ $\displaystyle = -\frac{1}{u}+C = -\frac{1}{g(x)}+C$
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