Question and Answers Forum

All Questions      Topic List

Integration Questions

Previous in All Question      Next in All Question      

Previous in Integration      Next in Integration      

Question Number 37258 by rahul 19 last updated on 11/Jun/18

∫ ((x^3 +1)/(√(x^2 +x))) dx = ?

$$\int\:\frac{{x}^{\mathrm{3}} +\mathrm{1}}{\sqrt{{x}^{\mathrm{2}} +{x}}}\:{dx}\:=\:? \\ $$

Answered by tanmay.chaudhury50@gmail.com last updated on 11/Jun/18

t^2 =x^2 +x 2tdt=(2x+1)dx ∫(((1/2)(2x+1−1)(x^2 +x−x)+1)/(√(x^2 +x)))dx ∫(1/2)×(((2x+1)(x^2 +x)−x(2x+1)−(x^2 +x)+x+1)/(√(x^2 +x))) =(1/2)∫(2x+1)(√(x^2 +x_ )) +((−2x^2 −x−x^2 −x+x+1)/(√(x^2 +x))) dx =(1/2)(2x+1)(√(x^2 +x)) +(1/2)∫((−3x^2 −x+1)/(√(x^2 +x)))dx =(1/2)∫(2x+1)(√(x^2 +_ x))+(1/2)∫((−3x^2 −3x+2x+1)/(√(x^2 +x))) =(1/2)∫(2x+1)(√(x^2 +x)) −(3/2)∫(√(x^2 +x)) − +(1/2)∫((2x+1)/(√(x^z +x)))dx =(1/(2 ))∫2tdt×t−(3/2)∫(√((x+(1/2))^2 −(1/4) )) +(1/2)∫((2tdt)/t) =(t^3 /3)−use formula+t =(((x^2 +x)^(3/2) )/3)+(√(x^2 +x)) −(3/2)∫(√((x+(1/2))^2 −(_ (1/4))))

$${t}^{\mathrm{2}} ={x}^{\mathrm{2}} +{x} \\ $$$$\mathrm{2}{tdt}=\left(\mathrm{2}{x}+\mathrm{1}\right){dx} \\ $$$$\int\frac{\frac{\mathrm{1}}{\mathrm{2}}\left(\mathrm{2}{x}+\mathrm{1}−\mathrm{1}\right)\left({x}^{\mathrm{2}} +{x}−{x}\right)+\mathrm{1}}{\sqrt{{x}^{\mathrm{2}} +{x}}}{dx} \\ $$$$\int\frac{\mathrm{1}}{\mathrm{2}}×\frac{\left(\mathrm{2}{x}+\mathrm{1}\right)\left({x}^{\mathrm{2}} +{x}\right)−{x}\left(\mathrm{2}{x}+\mathrm{1}\right)−\left({x}^{\mathrm{2}} +{x}\right)+{x}+\mathrm{1}}{\sqrt{{x}^{\mathrm{2}} +{x}}} \\ $$$$=\frac{\mathrm{1}}{\mathrm{2}}\int\left(\mathrm{2}{x}+\mathrm{1}\right)\sqrt{{x}^{\mathrm{2}} +{x}_{} }\:\:+\frac{−\mathrm{2}{x}^{\mathrm{2}} −{x}−{x}^{\mathrm{2}} −{x}+{x}+\mathrm{1}}{\sqrt{{x}^{\mathrm{2}} +{x}}}\:{dx} \\ $$$$=\frac{\mathrm{1}}{\mathrm{2}}\left(\mathrm{2}{x}+\mathrm{1}\right)\sqrt{{x}^{\mathrm{2}} +{x}}\:+\frac{\mathrm{1}}{\mathrm{2}}\int\frac{−\mathrm{3}{x}^{\mathrm{2}} −{x}+\mathrm{1}}{\sqrt{{x}^{\mathrm{2}} +{x}}}{dx} \\ $$$$=\frac{\mathrm{1}}{\mathrm{2}}\int\left(\mathrm{2}{x}+\mathrm{1}\right)\sqrt{{x}^{\mathrm{2}} +_{} {x}}+\frac{\mathrm{1}}{\mathrm{2}}\int\frac{−\mathrm{3}{x}^{\mathrm{2}} −\mathrm{3}{x}+\mathrm{2}{x}+\mathrm{1}}{\sqrt{{x}^{\mathrm{2}} +{x}}} \\ $$$$=\frac{\mathrm{1}}{\mathrm{2}}\int\left(\mathrm{2}{x}+\mathrm{1}\right)\sqrt{{x}^{\mathrm{2}} +{x}}\:−\frac{\mathrm{3}}{\mathrm{2}}\int\sqrt{{x}^{\mathrm{2}} +{x}}\:\:− \\ $$$$+\frac{\mathrm{1}}{\mathrm{2}}\int\frac{\mathrm{2}{x}+\mathrm{1}}{\sqrt{{x}^{{z}} +{x}}}{dx} \\ $$$$=\frac{\mathrm{1}}{\mathrm{2}\:}\int\mathrm{2}{tdt}×{t}−\frac{\mathrm{3}}{\mathrm{2}}\int\sqrt{\left({x}+\frac{\mathrm{1}}{\mathrm{2}}\right)^{\mathrm{2}} −\frac{\mathrm{1}}{\mathrm{4}}\:\:} \\ $$$$+\frac{\mathrm{1}}{\mathrm{2}}\int\frac{\mathrm{2}{tdt}}{{t}} \\ $$$$=\frac{{t}^{\mathrm{3}} }{\mathrm{3}}−{use}\:{formula}+{t} \\ $$$$=\frac{\left({x}^{\mathrm{2}} +{x}\right)^{\frac{\mathrm{3}}{\mathrm{2}}} }{\mathrm{3}}+\sqrt{{x}^{\mathrm{2}} +{x}}\:\:−\frac{\mathrm{3}}{\mathrm{2}}\int\sqrt{\left({x}+\frac{\mathrm{1}}{\mathrm{2}}\right)^{\mathrm{2}} −\left(_{} \frac{\mathrm{1}}{\mathrm{4}}\right)} \\ $$

Commented by rahul 19 last updated on 13/Jun/18

thank you sir.

$$\mathrm{thank}\:\mathrm{you}\:\mathrm{sir}. \\ $$

Terms of Service

Privacy Policy

Contact: info@tinkutara.com