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Category: Integration

find-the-area-abounded-y-x-afind-y-x-2-

Question Number 67106 by mhmd last updated on 22/Aug/19 $${find}\:{the}\:{area}\:{abounded}\:{y}=\sqrt{{x}} \\ $$$${afind}\:{y}={x}−\mathrm{2}? \\ $$ Commented by kaivan.ahmadi last updated on 24/Aug/19 $${x}−\mathrm{2}=\sqrt{{x}}\Rightarrow{x}^{\mathrm{2}} −\mathrm{5}{x}+\mathrm{4}=\mathrm{0}\Rightarrow\left({x}−\mathrm{1}\right)\left({x}−\mathrm{4}\right)=\mathrm{0}\Rightarrow \\ $$$$\begin{cases}{{x}=\mathrm{1}}\\{{x}=\mathrm{2}}\end{cases}…

sin-2-x-1-sin-2-x-dx-

Question Number 132610 by liberty last updated on 15/Feb/21 $$\Omega=\int\:\frac{\mathrm{sin}\:^{\mathrm{2}} \left(\mathrm{x}\right)}{\mathrm{1}+\mathrm{sin}\:^{\mathrm{2}} \left(\mathrm{x}\right)}\:\mathrm{dx}\: \\ $$ Answered by Dwaipayan Shikari last updated on 15/Feb/21 $$\int\frac{{sin}^{\mathrm{2}} {x}}{\mathrm{1}+{sin}^{\mathrm{2}} {x}}{dx}={x}−\int\frac{\mathrm{1}}{\mathrm{1}+{sin}^{\mathrm{2}}…

Question-67070

Question Number 67070 by mRDv143 last updated on 22/Aug/19 Commented by Prithwish sen last updated on 23/Aug/19 $$\int\frac{\mathrm{x}^{\mathrm{2}} −\mathrm{1}}{\left(\mathrm{x}+\mathrm{1}\right)^{\mathrm{2}} \:\sqrt{\mathrm{x}^{\mathrm{3}} +\mathrm{x}^{\mathrm{2}} +\mathrm{x}}}\:\mathrm{dx}=\int\frac{\mathrm{x}^{\mathrm{2}} −\mathrm{1}}{\left(\mathrm{x}+\mathrm{1}\right)^{\mathrm{2}} .\mathrm{x}\:\sqrt{\mathrm{x}+\frac{\mathrm{1}}{\mathrm{x}}+\mathrm{1}}}\:\mathrm{dx} \\…

Question-67069

Question Number 67069 by mRDv143 last updated on 22/Aug/19 Commented by Prithwish sen last updated on 22/Aug/19 $$\int\frac{\left(\mathrm{1}+\mathrm{logx}\right)^{\mathrm{2}} }{\mathrm{1}+\mathrm{xlogx}+\mathrm{logx}+\mathrm{x}\left(\mathrm{logx}\right)^{\mathrm{2}} }\:\mathrm{dx} \\ $$$$=\:\int\frac{\left(\mathrm{1}+\mathrm{logx}\right)^{\mathrm{2}} }{\left(\mathrm{1}+\mathrm{logx}\right)\left(\mathrm{1}+\mathrm{xlogx}\right)}\:\mathrm{dx} \\ $$$$=\int\frac{\mathrm{1}+\mathrm{logx}}{\mathrm{1}+\mathrm{xlogx}}\:\mathrm{dx}\:\:\mathrm{put}\:\mathrm{1}+\mathrm{xlogx}\:=\:\mathrm{u}\Rightarrow\left(\mathrm{1}+\mathrm{logx}\right)\mathrm{dx}=\:\mathrm{du}…