Menu Close

If-x-5-x-7-x-6-dx-p-ln-x-q-x-q-1-C-find-the-value-of-p-and-q-

Tinku Tara 0 Comments
Pin




Question Number 118230 by bobhans last updated on 16/Oct/20
If ∫ ((((√x))^5 )/(((√x))^7 +x^6 )) dx = p ln ((x^q /(x^q +1))) + C find the value of p and q.
$$\mathrm{If}\:\int\:\frac{\left(\sqrt{\mathrm{x}}\right)^{\mathrm{5}} }{\left(\sqrt{\mathrm{x}}\right)^{\mathrm{7}} +\mathrm{x}^{\mathrm{6}} }\:\mathrm{dx}\:=\:\mathrm{p}\:\mathrm{ln}\:\left(\frac{\mathrm{x}^{\mathrm{q}} }{\mathrm{x}^{\mathrm{q}} +\mathrm{1}}\right)\:+\:\mathrm{C}\: \\ $$$$\mathrm{find}\:\mathrm{the}\:\mathrm{value}\:\mathrm{of}\:\mathrm{p}\:\mathrm{and}\:\mathrm{q}. \\ $$
Answered by bemath last updated on 16/Oct/20
I=∫ ((x^2 (√x))/(x^3 (√x)+x^6 )) dx = ∫ ((√x)/(x(√x)+x^4 )) dx I= ∫ (dx/(x+x^3 (√x))) . let (√x) = u → dx = 2(√x) du I= ∫ ((2u du)/(u^2 +u^7 )) = ∫ ((2 du)/(u+u^6 )) I=∫ ((2 du)/(u(1+u^5 )))= ∫(( 2(1+u^5 −u^5 ))/(u(1+u^5 )))du =2∫ ((1/u)−(u^4 /(1+u^5 )))du = 2(ln u−(1/5)∫ ((d(1+u^5 ))/(1+u^5 ))) =2(ln u−(1/5)ln (1+u^5 ))+c = (2/5)(5ln u−ln (1+u^5 ))+c =(2/5) ln ((u^5 /(1+u^5 ))) +c = p ln ((x^q /(1+x^q )))+c =(2/5)ln ((x^(5/2) /(1+x^(5/2) )))+c = p ln ((x^q /(1+x^q )))+c we get { ((p=(2/5))),((q=(5/2))) :}
$${I}=\int\:\frac{{x}^{\mathrm{2}} \sqrt{{x}}}{{x}^{\mathrm{3}} \sqrt{{x}}+{x}^{\mathrm{6}} }\:{dx}\:=\:\int\:\frac{\sqrt{{x}}}{{x}\sqrt{{x}}+{x}^{\mathrm{4}} }\:{dx} \\ $$$${I}=\:\int\:\frac{{dx}}{{x}+{x}^{\mathrm{3}} \sqrt{{x}}}\:. \\ $$$${let}\:\sqrt{{x}}\:=\:{u}\:\rightarrow\:{dx}\:=\:\mathrm{2}\sqrt{{x}}\:{du} \\ $$$${I}=\:\int\:\frac{\mathrm{2}{u}\:{du}}{{u}^{\mathrm{2}} +{u}^{\mathrm{7}} }\:=\:\int\:\frac{\mathrm{2}\:{du}}{{u}+{u}^{\mathrm{6}} } \\ $$$${I}=\int\:\frac{\mathrm{2}\:{du}}{{u}\left(\mathrm{1}+{u}^{\mathrm{5}} \right)}=\:\int\frac{\:\mathrm{2}\left(\mathrm{1}+{u}^{\mathrm{5}} −{u}^{\mathrm{5}} \right)}{{u}\left(\mathrm{1}+{u}^{\mathrm{5}} \right)}{du} \\ $$$$=\mathrm{2}\int\:\left(\frac{\mathrm{1}}{{u}}−\frac{{u}^{\mathrm{4}} }{\mathrm{1}+{u}^{\mathrm{5}} }\right){du} \\ $$$$=\:\mathrm{2}\left(\mathrm{ln}\:{u}−\frac{\mathrm{1}}{\mathrm{5}}\int\:\frac{{d}\left(\mathrm{1}+{u}^{\mathrm{5}} \right)}{\mathrm{1}+{u}^{\mathrm{5}} }\right) \\ $$$$=\mathrm{2}\left(\mathrm{ln}\:{u}−\frac{\mathrm{1}}{\mathrm{5}}\mathrm{ln}\:\left(\mathrm{1}+{u}^{\mathrm{5}} \right)\right)+{c} \\ $$$$=\:\frac{\mathrm{2}}{\mathrm{5}}\left(\mathrm{5ln}\:{u}−\mathrm{ln}\:\left(\mathrm{1}+{u}^{\mathrm{5}} \right)\right)+{c} \\ $$$$=\frac{\mathrm{2}}{\mathrm{5}}\:\mathrm{ln}\:\left(\frac{{u}^{\mathrm{5}} }{\mathrm{1}+{u}^{\mathrm{5}} }\right)\:+{c}\:=\:{p}\:\mathrm{ln}\:\left(\frac{{x}^{{q}} }{\mathrm{1}+{x}^{{q}} }\right)+{c} \\ $$$$=\frac{\mathrm{2}}{\mathrm{5}}\mathrm{ln}\:\left(\frac{{x}^{\frac{\mathrm{5}}{\mathrm{2}}} }{\mathrm{1}+{x}^{\frac{\mathrm{5}}{\mathrm{2}}} }\right)+{c}\:=\:{p}\:\mathrm{ln}\:\left(\frac{{x}^{{q}} }{\mathrm{1}+{x}^{{q}} }\right)+{c}\: \\ $$$${we}\:{get}\:\begin{cases}{{p}=\frac{\mathrm{2}}{\mathrm{5}}}\\{{q}=\frac{\mathrm{5}}{\mathrm{2}}}\end{cases} \\ $$
Commented by bobhans last updated on 16/Oct/20
correct !
$${correct}\:! \\ $$

Pin

Leave a Reply

Your email address will not be published. Required fields are marked *