Question Number 5264 by Kasih last updated on 03/May/16
$${Prove}\:{that}\:\int\:\frac{\mathrm{2}{g}\left({x}\right){f}'\left({x}\right)−{f}\left({x}\right){g}'\left({x}\right)}{\mathrm{2}\left({g}\left({x}\right)\right)^{\frac{\mathrm{3}}{\mathrm{2}}} }\:{dx}\:=\frac{{f}\left({x}\right)}{\:\sqrt{{g}\left({x}\right)}}\:+\:{C} \\ $$
Answered by Yozzii last updated on 03/May/16
![Let I=∫((2g(x)f^′ (x)−f(x)g^′ (x))/(2(g(x))^(3/2) ))dx. I=∫[f^′ (x)(g(x))^(−1/2) +f(x){−(1/2)g^′ (x)(g(x))^(−3/2) }]dx I=∫[(d/dx){f(x)}(g(x))^(−1/2) +f(x)(d/dx){(g(x))^(−1/2) }]dx I=∫(d/dx){f(x)(g(x))^(−1/2) }dx The line above says to take the derivative of the function f(x)(g(x))^(−1/2) w.r.t x, and then take the antiderivative of this result w.r.t x. This hence gives back the function f(x)(g(x))^(−1/2) but possibly with an added constant C since (d/dx)(f(x)(g(x))^(−1/2) +C)=(d/dx)(f(x)(g(x))^(−1/2) ). ⇒I=f(x)(g(x))^(−1/2) +C as the required general result.](https://www.tinkutara.com/question/Q5269.png)
$${Let}\:{I}=\int\frac{\mathrm{2}{g}\left({x}\right){f}^{'} \left({x}\right)−{f}\left({x}\right){g}^{'} \left({x}\right)}{\mathrm{2}\left({g}\left({x}\right)\right)^{\mathrm{3}/\mathrm{2}} }{dx}. \\ $$$${I}=\int\left[{f}^{'} \left({x}\right)\left({g}\left({x}\right)\right)^{−\mathrm{1}/\mathrm{2}} +{f}\left({x}\right)\left\{−\frac{\mathrm{1}}{\mathrm{2}}{g}^{'} \left({x}\right)\left({g}\left({x}\right)\right)^{−\mathrm{3}/\mathrm{2}} \right\}\right]{dx} \\ $$$${I}=\int\left[\frac{{d}}{{dx}}\left\{{f}\left({x}\right)\right\}\left({g}\left({x}\right)\right)^{−\mathrm{1}/\mathrm{2}} +{f}\left({x}\right)\frac{{d}}{{dx}}\left\{\left({g}\left({x}\right)\right)^{−\mathrm{1}/\mathrm{2}} \right\}\right]{dx} \\ $$$${I}=\int\frac{{d}}{{dx}}\left\{{f}\left({x}\right)\left({g}\left({x}\right)\right)^{−\mathrm{1}/\mathrm{2}} \right\}{dx} \\ $$$${The}\:{line}\:{above}\:{says}\:{to}\:{take}\:{the}\:{derivative} \\ $$$${of}\:{the}\:{function}\:{f}\left({x}\right)\left({g}\left({x}\right)\right)^{−\mathrm{1}/\mathrm{2}} \:{w}.{r}.{t}\:{x},\:{and}\:{then} \\ $$$${take}\:{the}\:{antiderivative}\:{of}\:{this}\:{result}\:{w}.{r}.{t}\:{x}. \\ $$$${This}\:{hence}\:{gives}\:{back}\:{the}\:{function} \\ $$$${f}\left({x}\right)\left({g}\left({x}\right)\right)^{−\mathrm{1}/\mathrm{2}} \:{but}\:{possibly}\:{with}\:{an}\: \\ $$$${added}\:{constant}\:{C}\:{since}\:\frac{{d}}{{dx}}\left({f}\left({x}\right)\left({g}\left({x}\right)\right)^{−\mathrm{1}/\mathrm{2}} +{C}\right)=\frac{{d}}{{dx}}\left({f}\left({x}\right)\left({g}\left({x}\right)\right)^{−\mathrm{1}/\mathrm{2}} \right). \\ $$$$\Rightarrow{I}={f}\left({x}\right)\left({g}\left({x}\right)\right)^{−\mathrm{1}/\mathrm{2}} +{C}\:{as}\:{the}\:{required} \\ $$$${general}\:{result}. \\ $$$$ \\ $$