Question Number 39020 by maxmathsup by imad last updated on 01/Jul/18
$${calculate}\:\:\int_{\mathrm{0}} ^{\mathrm{1}} \:\:\frac{{ln}\left(\mathrm{1}+\sqrt{{x}^{\mathrm{2}} \:+\mathrm{1}}\right)}{\sqrt{{x}^{\mathrm{2}} \:+\mathrm{1}}}\:{dx} \\ $$
Answered by behi83417@gmail.com last updated on 02/Jul/18
![x=tgt⇒dx=(1+tg^2 t)dt I=∫((ln(1+sect))/(sect)).sec^2 tdt= =∫sect.ln(1+sect)dt= sect.ln(1+sect)−∫sect.((sint.sec^2 t)/(1+sect))dt= =do−∫((sint.(1/(cos^3 t)))/(1+(1/(cost))))dt=do−∫((sintdt)/(cos^3 t+cos^2 t)) cost=u⇒∫((−sintdt)/(cos^3 t+cos^2 t))=∫(du/(u^3 +u^2 ))= =∫(du/(u^2 (u+1)))=∫[((−1)/u)+(1/u^2 )+(1/(u+1))]du =−lnu−u+ln(u+1)+c= =−lncost−cost+ln(1+cost)+c= =ln((1+cost)/(cost))−cost+c=ln(1+sect)−cost+c ⇒I=(√(1+x^2 )).ln(1+(√(1+x^2 )))+ln(1+(√(1+x^2 )))−(1/(√(1+x^2 )))+c= =(1+(√(1+x^2 )))ln(1+(√(1+x^2 )))−(1/(√(1+x^2 )))+c.■](Q39052.png)
$${x}={tgt}\Rightarrow{dx}=\left(\mathrm{1}+{tg}^{\mathrm{2}} {t}\right){dt} \\ $$$${I}=\int\frac{{ln}\left(\mathrm{1}+{sect}\right)}{{sect}}.{sec}^{\mathrm{2}} {tdt}= \\ $$$$=\int{sect}.{ln}\left(\mathrm{1}+{sect}\right){dt}= \\ $$$${sect}.{ln}\left(\mathrm{1}+{sect}\right)−\int{sect}.\frac{{sint}.{sec}^{\mathrm{2}} {t}}{\mathrm{1}+{sect}}{dt}= \\ $$$$={do}−\int\frac{{sint}.\frac{\mathrm{1}}{{cos}^{\mathrm{3}} {t}}}{\mathrm{1}+\frac{\mathrm{1}}{{cost}}}{dt}={do}−\int\frac{{sintdt}}{{cos}^{\mathrm{3}} {t}+{cos}^{\mathrm{2}} {t}} \\ $$$${cost}={u}\Rightarrow\int\frac{−{sintdt}}{{cos}^{\mathrm{3}} {t}+{cos}^{\mathrm{2}} {t}}=\int\frac{{du}}{{u}^{\mathrm{3}} +{u}^{\mathrm{2}} }= \\ $$$$=\int\frac{{du}}{{u}^{\mathrm{2}} \left({u}+\mathrm{1}\right)}=\int\left[\frac{−\mathrm{1}}{{u}}+\frac{\mathrm{1}}{{u}^{\mathrm{2}} }+\frac{\mathrm{1}}{{u}+\mathrm{1}}\right]{du} \\ $$$$=−{lnu}−{u}+{ln}\left({u}+\mathrm{1}\right)+{c}= \\ $$$$=−{lncost}−{cost}+{ln}\left(\mathrm{1}+{cost}\right)+{c}= \\ $$$$={ln}\frac{\mathrm{1}+{cost}}{{cost}}−{cost}+{c}={ln}\left(\mathrm{1}+{sect}\right)−{cost}+{c} \\ $$$$\Rightarrow{I}=\sqrt{\mathrm{1}+\boldsymbol{{x}}^{\mathrm{2}} }.\boldsymbol{{ln}}\left(\mathrm{1}+\sqrt{\mathrm{1}+\boldsymbol{{x}}^{\mathrm{2}} }\right)+\boldsymbol{{ln}}\left(\mathrm{1}+\sqrt{\mathrm{1}+\boldsymbol{{x}}^{\mathrm{2}} }\right)−\frac{\mathrm{1}}{\sqrt{\mathrm{1}+\boldsymbol{{x}}^{\mathrm{2}} }}+\boldsymbol{{c}}= \\ $$$$=\left(\mathrm{1}+\sqrt{\mathrm{1}+\boldsymbol{{x}}^{\mathrm{2}} }\right)\boldsymbol{{ln}}\left(\mathrm{1}+\sqrt{\mathrm{1}+\boldsymbol{{x}}^{\mathrm{2}} }\right)−\frac{\mathrm{1}}{\sqrt{\mathrm{1}+\boldsymbol{{x}}^{\mathrm{2}} }}+\boldsymbol{{c}}.\blacksquare \\ $$