Question Number 57433 by naka3546 last updated on 04/Apr/19
Answered by tanmay.chaudhury50@gmail.com last updated on 04/Apr/19
$$\int\frac{{x}^{\mathrm{4}} +\mathrm{1}}{{x}^{\mathrm{2}} \sqrt{{x}^{\mathrm{2}} \left({x}^{\mathrm{2}} −\frac{\mathrm{1}}{{x}^{\mathrm{2}} }\right)}}{dx} \\ $$$$\int\frac{{x}+\frac{\mathrm{1}}{{x}^{\mathrm{3}} }}{\:\sqrt{{x}^{\mathrm{2}} −\frac{\mathrm{1}}{{x}^{\mathrm{2}} }}}{dx} \\ $$$${t}^{\mathrm{2}} ={x}^{\mathrm{2}} −\frac{\mathrm{1}}{{x}^{\mathrm{2}} } \\ $$$$\mathrm{2}{tdt}=\left(\mathrm{2}{x}+\frac{\mathrm{2}}{{x}^{\mathrm{3}} }\right){dx} \\ $$$${tdt}=\left({x}+\frac{\mathrm{1}}{{x}^{\mathrm{3}} }\right){dx} \\ $$$$\int\frac{{tdt}}{{t}} \\ $$$$={t}+{c} \\ $$$$=\sqrt{{x}^{\mathrm{2}} −\frac{\mathrm{1}}{{x}^{\mathrm{2}} }\:}\:\:+{c} \\ $$
Commented by tanmay.chaudhury50@gmail.com last updated on 04/Apr/19
$${pls}\:{check}\:…{tricky}\:{problem}… \\ $$
Commented by MJS last updated on 04/Apr/19
$$\mathrm{your}\:\mathrm{solution}\:\mathrm{is}\:\mathrm{correct} \\ $$
Commented by tanmay.chaudhury50@gmail.com last updated on 05/Apr/19
$${thank}\:{you}\:{sir}… \\ $$
Commented by malwaan last updated on 05/Apr/19
$$\boldsymbol{{and}}\:\sqrt{\boldsymbol{{x}}^{\mathrm{2}} −\frac{\mathrm{1}}{\boldsymbol{{x}}^{\mathrm{2}} }}\:=\frac{\sqrt{\boldsymbol{{x}}^{\mathrm{4}} −\mathrm{1}}}{\boldsymbol{{x}}}\: \\ $$$$\boldsymbol{{thank}}\:\boldsymbol{{you}}\:\boldsymbol{{so}}\:\boldsymbol{{much}} \\ $$
Answered by MJS last updated on 04/Apr/19
$$\mathrm{possible}\:\mathrm{hat}\:\mathrm{trick}: \\ $$$$\int\frac{{x}^{\mathrm{4}} +\mathrm{1}}{{x}^{\mathrm{2}} \sqrt{{x}^{\mathrm{4}} −\mathrm{1}}}{dx}=\int\frac{\mathrm{2}{x}^{\mathrm{2}} }{\:\sqrt{{x}^{\mathrm{4}} −\mathrm{1}}}{dx}−\int\frac{\sqrt{{x}^{\mathrm{4}} −\mathrm{1}}}{{x}^{\mathrm{2}} }{dx} \\ $$$$\mathrm{now}\:\mathrm{write}\:\int\frac{\sqrt{{x}^{\mathrm{4}} −\mathrm{1}}}{{x}^{\mathrm{2}} }{dx}\:\mathrm{by}\:\mathrm{parts}\:\mathrm{without} \\ $$$$\mathrm{solving}: \\ $$$$\int{u}'{v}={uv}−\int{uv}' \\ $$$${u}'=\frac{\mathrm{1}}{{x}^{\mathrm{2}} }\:\Rightarrow\:{u}=−\frac{\mathrm{1}}{{x}} \\ $$$${v}=\sqrt{{x}^{\mathrm{4}} −\mathrm{1}}\:\Rightarrow\:{v}'=\frac{\mathrm{2}{x}^{\mathrm{3}} }{\:\sqrt{{x}^{\mathrm{4}} −\mathrm{1}}} \\ $$$$\mathrm{so}\:\mathrm{we}\:\mathrm{have}: \\ $$$$\int\frac{{x}^{\mathrm{4}} +\mathrm{1}}{{x}^{\mathrm{2}} \sqrt{{x}^{\mathrm{4}} −\mathrm{1}}}{dx}=\int\frac{\mathrm{2}{x}^{\mathrm{2}} }{\:\sqrt{{x}^{\mathrm{4}} −\mathrm{1}}}{dx}−\left(−\frac{\sqrt{{x}^{\mathrm{4}} −\mathrm{1}}}{{x}}−\int−\frac{\mathrm{2}{x}^{\mathrm{2}} }{\:\sqrt{{x}^{\mathrm{4}} −\mathrm{1}}}{dx}\right)= \\ $$$$=\frac{\sqrt{{x}^{\mathrm{4}} −\mathrm{1}}}{{x}}+{C} \\ $$