Question and Answers Forum

All Questions      Topic List

Arithmetic Questions

Previous in All Question      Next in All Question      

Previous in Arithmetic      Next in Arithmetic      

Question Number 121066 by Jamshidbek2311 last updated on 05/Nov/20

1)∫(x/(x+1))dx=? 2)∫(x/(√(x−1)))dx=? 3)∫x(√(x−4))dx=?^

$$\left.\mathrm{1}\right)\int\frac{{x}}{{x}+\mathrm{1}}{dx}=? \\ $$$$\left.\mathrm{2}\right)\int\frac{{x}}{\sqrt{{x}−\mathrm{1}}}{dx}=? \\ $$$$\left.\mathrm{3}\right)\int{x}\sqrt{{x}−\mathrm{4}}{dx}=\overset{} {?} \\ $$$$ \\ $$

Commented by liberty last updated on 05/Nov/20

1) ∫ ((x+1−1)/(x+1)) dx = ∫1−(1/(x+1)) dx =x−ln ∣x+1∣ + c = ln e^x −ln ∣x+1∣ + c = ln ∣(e^x /(x+1)) ∣ + c

$$\left.\mathrm{1}\right)\:\int\:\frac{\mathrm{x}+\mathrm{1}−\mathrm{1}}{\mathrm{x}+\mathrm{1}}\:\mathrm{dx}\:=\:\int\mathrm{1}−\frac{\mathrm{1}}{\mathrm{x}+\mathrm{1}}\:\mathrm{dx} \\ $$$$=\mathrm{x}−\mathrm{ln}\:\mid\mathrm{x}+\mathrm{1}\mid\:+\:\mathrm{c}\:=\:\mathrm{ln}\:\mathrm{e}^{\mathrm{x}} −\mathrm{ln}\:\mid\mathrm{x}+\mathrm{1}\mid\:+\:\mathrm{c} \\ $$$$=\:\mathrm{ln}\:\mid\frac{\mathrm{e}^{\mathrm{x}} }{\mathrm{x}+\mathrm{1}}\:\mid\:+\:\mathrm{c}\: \\ $$

Answered by Dwaipayan Shikari last updated on 05/Nov/20

∫(x/( (√(x−1))))dx =∫((2tdt)/t)(t^2 +1) (x−1)=t =((2t^3 )/3)+2t=(2/3)(x−1)^(3/2) +2(√(x−1)) +C

$$\int\frac{{x}}{\:\sqrt{{x}−\mathrm{1}}}{dx} \\ $$$$=\int\frac{\mathrm{2}{tdt}}{{t}}\left({t}^{\mathrm{2}} +\mathrm{1}\right)\:\:\:\:\:\:\:\:\:\:\left({x}−\mathrm{1}\right)={t} \\ $$$$=\frac{\mathrm{2}{t}^{\mathrm{3}} }{\mathrm{3}}+\mathrm{2}{t}=\frac{\mathrm{2}}{\mathrm{3}}\left({x}−\mathrm{1}\right)^{\frac{\mathrm{3}}{\mathrm{2}}} +\mathrm{2}\sqrt{{x}−\mathrm{1}}\:+{C} \\ $$

Answered by bramlexs22 last updated on 05/Nov/20

3) ∫ x(√(x−4)) dx set (√(x−4)) = z →x = z^2 +4 ∫ (z^2 +4)z(2z dz)= ∫ (z^2 +4)(2z^2 ) dz = ∫ 2z^4 +8z^2 dz = (2/5)z^5 +(8/3)z^3 +c = z^3 ((2/5)z^2 +(8/3))+c = (x−4)^(3/2) ((2/5)(x−4)+(8/3))+c

$$\left.\mathrm{3}\right)\:\int\:\mathrm{x}\sqrt{\mathrm{x}−\mathrm{4}}\:\mathrm{dx}\: \\ $$$$\:\mathrm{set}\:\sqrt{\mathrm{x}−\mathrm{4}}\:=\:\mathrm{z}\:\rightarrow\mathrm{x}\:=\:\mathrm{z}^{\mathrm{2}} +\mathrm{4} \\ $$$$\int\:\left(\mathrm{z}^{\mathrm{2}} +\mathrm{4}\right)\mathrm{z}\left(\mathrm{2z}\:\mathrm{dz}\right)= \\ $$$$\int\:\left(\mathrm{z}^{\mathrm{2}} +\mathrm{4}\right)\left(\mathrm{2z}^{\mathrm{2}} \right)\:\mathrm{dz}\:= \\ $$$$\int\:\mathrm{2z}^{\mathrm{4}} +\mathrm{8z}^{\mathrm{2}} \:\mathrm{dz}\:=\:\frac{\mathrm{2}}{\mathrm{5}}\mathrm{z}^{\mathrm{5}} +\frac{\mathrm{8}}{\mathrm{3}}\mathrm{z}^{\mathrm{3}} +\mathrm{c} \\ $$$$=\:\mathrm{z}^{\mathrm{3}} \left(\frac{\mathrm{2}}{\mathrm{5}}\mathrm{z}^{\mathrm{2}} +\frac{\mathrm{8}}{\mathrm{3}}\right)+\mathrm{c} \\ $$$$=\:\left(\mathrm{x}−\mathrm{4}\right)^{\frac{\mathrm{3}}{\mathrm{2}}} \left(\frac{\mathrm{2}}{\mathrm{5}}\left(\mathrm{x}−\mathrm{4}\right)+\frac{\mathrm{8}}{\mathrm{3}}\right)+\mathrm{c} \\ $$

Answered by Dwaipayan Shikari last updated on 05/Nov/20

∫x(√(x−4)) dx =∫2t(t^2 +4)tdt t^2 =x−4 =(2/5)t^5 +(8/3)t^3 =(2/5)(x−4)^(5/2) +(8/3)(x−4)^(3/2)

$$\int{x}\sqrt{{x}−\mathrm{4}}\:{dx} \\ $$$$=\int\mathrm{2}{t}\left({t}^{\mathrm{2}} +\mathrm{4}\right){tdt}\:\:\:\:\:\:\:\:\:{t}^{\mathrm{2}} ={x}−\mathrm{4} \\ $$$$=\frac{\mathrm{2}}{\mathrm{5}}{t}^{\mathrm{5}} +\frac{\mathrm{8}}{\mathrm{3}}{t}^{\mathrm{3}} =\frac{\mathrm{2}}{\mathrm{5}}\left({x}−\mathrm{4}\right)^{\frac{\mathrm{5}}{\mathrm{2}}} +\frac{\mathrm{8}}{\mathrm{3}}\left({x}−\mathrm{4}\right)^{\frac{\mathrm{3}}{\mathrm{2}}} \\ $$

Terms of Service

Privacy Policy

Contact: info@tinkutara.com