Question and Answers Forum

All Questions      Topic List

Integration Questions

Previous in All Question      Next in All Question      

Previous in Integration      Next in Integration      

Question Number 5264 by Kasih last updated on 03/May/16

Prove that ∫ ((2g(x)f′(x)−f(x)g′(x))/(2(g(x))^(3/2) )) dx =((f(x))/(√(g(x)))) + C

$${Prove}\:{that}\:\int\:\frac{\mathrm{2}{g}\left({x}\right){f}'\left({x}\right)−{f}\left({x}\right){g}'\left({x}\right)}{\mathrm{2}\left({g}\left({x}\right)\right)^{\frac{\mathrm{3}}{\mathrm{2}}} }\:{dx}\:=\frac{{f}\left({x}\right)}{\sqrt{{g}\left({x}\right)}}\:+\:{C} \\ $$

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.

$${Let}\:{I}=\int\frac{\mathrm{2}{g}\left({x}\right){f}^{'} \left({x}\right)−{f}\left({x}\right){g}^{'} \left({x}\right)}{\mathrm{2}\left({g}\left({x}\right)\right)^{\mathrm{3}/\mathrm{2}} }{dx}. \\ $$$${I}=\int\left[{f}^{'} \left({x}\right)\left({g}\left({x}\right)\right)^{−\mathrm{1}/\mathrm{2}} +{f}\left({x}\right)\left\{−\frac{\mathrm{1}}{\mathrm{2}}{g}^{'} \left({x}\right)\left({g}\left({x}\right)\right)^{−\mathrm{3}/\mathrm{2}} \right\}\right]{dx} \\ $$$${I}=\int\left[\frac{{d}}{{dx}}\left\{{f}\left({x}\right)\right\}\left({g}\left({x}\right)\right)^{−\mathrm{1}/\mathrm{2}} +{f}\left({x}\right)\frac{{d}}{{dx}}\left\{\left({g}\left({x}\right)\right)^{−\mathrm{1}/\mathrm{2}} \right\}\right]{dx} \\ $$$${I}=\int\frac{{d}}{{dx}}\left\{{f}\left({x}\right)\left({g}\left({x}\right)\right)^{−\mathrm{1}/\mathrm{2}} \right\}{dx} \\ $$$${The}\:{line}\:{above}\:{says}\:{to}\:{take}\:{the}\:{derivative} \\ $$$${of}\:{the}\:{function}\:{f}\left({x}\right)\left({g}\left({x}\right)\right)^{−\mathrm{1}/\mathrm{2}} \:{w}.{r}.{t}\:{x},\:{and}\:{then} \\ $$$${take}\:{the}\:{antiderivative}\:{of}\:{this}\:{result}\:{w}.{r}.{t}\:{x}. \\ $$$${This}\:{hence}\:{gives}\:{back}\:{the}\:{function} \\ $$$${f}\left({x}\right)\left({g}\left({x}\right)\right)^{−\mathrm{1}/\mathrm{2}} \:{but}\:{possibly}\:{with}\:{an}\: \\ $$$${added}\:{constant}\:{C}\:{since}\:\frac{{d}}{{dx}}\left({f}\left({x}\right)\left({g}\left({x}\right)\right)^{−\mathrm{1}/\mathrm{2}} +{C}\right)=\frac{{d}}{{dx}}\left({f}\left({x}\right)\left({g}\left({x}\right)\right)^{−\mathrm{1}/\mathrm{2}} \right). \\ $$$$\Rightarrow{I}={f}\left({x}\right)\left({g}\left({x}\right)\right)^{−\mathrm{1}/\mathrm{2}} +{C}\:{as}\:{the}\:{required} \\ $$$${general}\:{result}. \\ $$$$ \\ $$

Terms of Service

Privacy Policy

Contact: info@tinkutara.com