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Category: Integration

2x-2-3x-3-x-1-x-2-2x-5-dx-

Question Number 125440 by bramlexs22 last updated on 11/Dec/20 $$\:\:\:\:\:\:\:\:\:\:\:\int\frac{\mathrm{2}{x}^{\mathrm{2}} −\mathrm{3}{x}−\mathrm{3}}{\left({x}−\mathrm{1}\right)\left({x}^{\mathrm{2}} −\mathrm{2}{x}+\mathrm{5}\right)}\:{dx}\: \\ $$ Answered by Ar Brandon last updated on 11/Dec/20 $$\mathrm{f}\left(\mathrm{x}\right)=\frac{\mathrm{2x}^{\mathrm{2}} −\mathrm{3x}−\mathrm{3}}{\left(\mathrm{x}−\mathrm{1}\right)\left(\mathrm{x}^{\mathrm{2}} −\mathrm{2x}+\mathrm{5}\right)}=\frac{\mathrm{a}}{\mathrm{x}−\mathrm{1}}+\frac{\mathrm{bx}+\mathrm{c}}{\mathrm{x}^{\mathrm{2}}…

2x-1-20-

Question Number 59905 by Sardor2211 last updated on 15/May/19 $$\int\left(\mathrm{2x}−\mathrm{1}\hat {\right)}\mathrm{20} \\ $$ Commented by maxmathsup by imad last updated on 16/May/19 $${if}\:{you}\:{mean}\:{I}\:=\int\:\left(\mathrm{2}{x}−\mathrm{1}\right)^{\mathrm{20}} {dx}\:\Rightarrow{I}\:=\int\:\sum_{{k}=\mathrm{0}} ^{\mathrm{20}}…

Question-59893

Question Number 59893 by aliesam last updated on 15/May/19 Commented by maxmathsup by imad last updated on 15/May/19 $${method}\:{using}\:{the}\:{formuae}\:\int_{\mathrm{0}} ^{\infty} \frac{{t}^{{a}−\mathrm{1}} }{\mathrm{1}+{t}}{dt}\:=\frac{\pi}{{sin}\left(\pi{a}\right)}\:\:{if}\:\mathrm{0}<{a}<\mathrm{1} \\ $$$${let}\:{I}\:=\int_{−\infty} ^{+\infty}…

0-100-dx-x-100-x-

Question Number 125421 by bemath last updated on 11/Dec/20 $$\:\underset{\mathrm{0}} {\overset{\mathrm{100}} {\int}}\:\frac{{dx}}{\:\sqrt{{x}\left(\mathrm{100}−{x}\right)}}\:?\: \\ $$ Answered by liberty last updated on 11/Dec/20 $${I}=\underset{\mathrm{0}} {\overset{\mathrm{100}} {\int}}\:\frac{{dx}}{\:\sqrt{{x}}\:\sqrt{\mathrm{100}−{x}}}\:;\:{let}\:\sqrt{{x}}\:=\:\mathrm{10}\:\mathrm{sin}\:{t}\: \\…

dx-x-x-x-2-

Question Number 125416 by bemath last updated on 10/Dec/20 $$\int\:\frac{{dx}}{\:\sqrt{{x}\sqrt{{x}}−{x}^{\mathrm{2}} }}\:? \\ $$ Answered by Dwaipayan Shikari last updated on 10/Dec/20 $$\int\frac{{dx}}{\:\sqrt{{x}}\sqrt{\sqrt{{x}}−{x}}}\:\:\:\:\:\:\:\:\:\:\sqrt{{x}}={t}\Rightarrow\frac{\mathrm{1}}{\mathrm{2}\sqrt{{x}}}=\frac{{dt}}{{dx}} \\ $$$$=\mathrm{2}\int\frac{{dt}}{\:\sqrt{{t}−{t}^{\mathrm{2}} }}=\mathrm{2}\int\frac{\mathrm{1}}{\:\sqrt{{t}}\left(\sqrt{\mathrm{1}−{t}}\right)}{dt}\:\:\:\:\:\:\:\:\:\:\:\:\:{t}={u}^{\mathrm{2}}…