Question Number 5266 by Kasih last updated on 03/May/16
$$\int\:\frac{\mathrm{3}{x}}{\:\sqrt{{x}^{\mathrm{2}} +\:\mathrm{2}{x}+\:\mathrm{5}}}\:{dx} \\ $$
Commented by prakash jain last updated on 03/May/16
$$\mathrm{You}\:\mathrm{can}\:\mathrm{integrate}\:\mathrm{as}\:\mathrm{following} \\ $$$$\frac{\mathrm{3}{x}+\mathrm{3}}{\:\sqrt{{x}^{\mathrm{2}} +\mathrm{2}{x}+\mathrm{5}}}−\frac{\mathrm{3}}{\:\sqrt{{x}^{\mathrm{2}} +\mathrm{2}{x}+\mathrm{5}}} \\ $$$$\mathrm{for}\:\mathrm{the}\:\mathrm{fist}\:\mathrm{part}\:\mathrm{Note}\:\mathrm{that}\:\frac{{d}}{{dx}}\left({x}^{\mathrm{2}} +\mathrm{2}{x}+\mathrm{5}\right)=\mathrm{2}\left({x}+\mathrm{1}\right) \\ $$$$\mathrm{The}\:\mathrm{second}\:\mathrm{part} \\ $$$${x}^{\mathrm{2}} +\mathrm{2}{x}+\mathrm{5}=\left({x}+\mathrm{1}\right)^{\mathrm{2}} +\mathrm{4} \\ $$$$\mathrm{integrate}\:\mathrm{by}\:\mathrm{using}\:\mathrm{formula}\:\mathrm{for} \\ $$$$\frac{\mathrm{1}}{\:\sqrt{{x}^{\mathrm{2}} +{a}^{\mathrm{2}} }} \\ $$
Answered by Yozzii last updated on 03/May/16
![Let I=∫((3x)/( (√(x^2 +2x+5))))dx. I=(3/2)∫((2x+2−2)/( (√(x^2 +2x+5))))dx I=(3/2)∫[(((d/dx)(x^2 +2x+5))/( (√(x^2 +2x+5))))−(2/( (√(4+(x+1)^2 ))))dx] Let u=x^2 +2x+5⇒du=(2x+2)dx ∴∫((2x+2)/( (√(x^2 +2x+5))))dx=∫(du/( (√u)))=2(√u)+κ=2(√(x^2 +2x+5))+κ Let x+1=2sinht⇒dx=2coshtdt. Also, t=sinh^(−1) (((x+1)/2))=ln(((x+1)/2)+(1/2)(√(4+(x+1)^2 )))=ln(x+1+(√(x^2 +2x+5)))−ln2 ∴4+(x+1)^2 =4+4sinh^2 t=4(1+sinh^2 t)=4cosh^2 t ∴∫(2/( (√(4+(x+1)^2 ))))dx=∫((2×2cosht)/( (√(4cosh^2 t))))dt=2∫dt=2t+ϑ=2ln(x+1+(√(x^2 +2x+5)))−2ln2+ϑ ∴I=(3/2)(2(√(x^2 +2x+5))+κ−2ln(x+1+(√(x^2 +2x+5)))+2ln2−ϑ) I=3(√(x^2 +2x+5))−3ln(x+1+(√(x^2 +2x+5)))+C where C=(3/2)κ+3ln2−(3/2)ϑ=constant](https://www.tinkutara.com/question/Q5272.png)
$${Let}\:{I}=\int\frac{\mathrm{3}{x}}{\:\sqrt{{x}^{\mathrm{2}} +\mathrm{2}{x}+\mathrm{5}}}{dx}. \\ $$$${I}=\frac{\mathrm{3}}{\mathrm{2}}\int\frac{\mathrm{2}{x}+\mathrm{2}−\mathrm{2}}{\:\sqrt{{x}^{\mathrm{2}} +\mathrm{2}{x}+\mathrm{5}}}{dx} \\ $$$${I}=\frac{\mathrm{3}}{\mathrm{2}}\int\left[\frac{\frac{{d}}{{dx}}\left({x}^{\mathrm{2}} +\mathrm{2}{x}+\mathrm{5}\right)}{\:\sqrt{{x}^{\mathrm{2}} +\mathrm{2}{x}+\mathrm{5}}}−\frac{\mathrm{2}}{\:\sqrt{\mathrm{4}+\left({x}+\mathrm{1}\right)^{\mathrm{2}} }}{dx}\right] \\ $$$${Let}\:{u}={x}^{\mathrm{2}} +\mathrm{2}{x}+\mathrm{5}\Rightarrow{du}=\left(\mathrm{2}{x}+\mathrm{2}\right){dx} \\ $$$$\therefore\int\frac{\mathrm{2}{x}+\mathrm{2}}{\:\sqrt{{x}^{\mathrm{2}} +\mathrm{2}{x}+\mathrm{5}}}{dx}=\int\frac{{du}}{\:\sqrt{{u}}}=\mathrm{2}\sqrt{{u}}+\kappa=\mathrm{2}\sqrt{{x}^{\mathrm{2}} +\mathrm{2}{x}+\mathrm{5}}+\kappa \\ $$$$ \\ $$$${Let}\:{x}+\mathrm{1}=\mathrm{2}{sinht}\Rightarrow{dx}=\mathrm{2}{coshtdt}. \\ $$$${Also},\:{t}={sinh}^{−\mathrm{1}} \left(\frac{{x}+\mathrm{1}}{\mathrm{2}}\right)={ln}\left(\frac{{x}+\mathrm{1}}{\mathrm{2}}+\frac{\mathrm{1}}{\mathrm{2}}\sqrt{\mathrm{4}+\left({x}+\mathrm{1}\right)^{\mathrm{2}} }\right)={ln}\left({x}+\mathrm{1}+\sqrt{{x}^{\mathrm{2}} +\mathrm{2}{x}+\mathrm{5}}\right)−{ln}\mathrm{2} \\ $$$$\therefore\mathrm{4}+\left({x}+\mathrm{1}\right)^{\mathrm{2}} =\mathrm{4}+\mathrm{4}{sinh}^{\mathrm{2}} {t}=\mathrm{4}\left(\mathrm{1}+{sinh}^{\mathrm{2}} {t}\right)=\mathrm{4}{cosh}^{\mathrm{2}} {t} \\ $$$$\therefore\int\frac{\mathrm{2}}{\:\sqrt{\mathrm{4}+\left({x}+\mathrm{1}\right)^{\mathrm{2}} }}{dx}=\int\frac{\mathrm{2}×\mathrm{2}{cosht}}{\:\sqrt{\mathrm{4}{cosh}^{\mathrm{2}} {t}}}{dt}=\mathrm{2}\int{dt}=\mathrm{2}{t}+\vartheta=\mathrm{2}{ln}\left({x}+\mathrm{1}+\sqrt{{x}^{\mathrm{2}} +\mathrm{2}{x}+\mathrm{5}}\right)−\mathrm{2}{ln}\mathrm{2}+\vartheta \\ $$$$ \\ $$$$\therefore{I}=\frac{\mathrm{3}}{\mathrm{2}}\left(\mathrm{2}\sqrt{{x}^{\mathrm{2}} +\mathrm{2}{x}+\mathrm{5}}+\kappa−\mathrm{2}{ln}\left({x}+\mathrm{1}+\sqrt{{x}^{\mathrm{2}} +\mathrm{2}{x}+\mathrm{5}}\right)+\mathrm{2}{ln}\mathrm{2}−\vartheta\right) \\ $$$${I}=\mathrm{3}\sqrt{{x}^{\mathrm{2}} +\mathrm{2}{x}+\mathrm{5}}−\mathrm{3}{ln}\left({x}+\mathrm{1}+\sqrt{{x}^{\mathrm{2}} +\mathrm{2}{x}+\mathrm{5}}\right)+{C} \\ $$$${where}\:{C}=\frac{\mathrm{3}}{\mathrm{2}}\kappa+\mathrm{3}{ln}\mathrm{2}−\frac{\mathrm{3}}{\mathrm{2}}\vartheta={constant} \\ $$$$ \\ $$$$ \\ $$