Question Number 110619 by mathdave last updated on 29/Aug/20
![someone posted this problem and my solution followed solve ∫_0 ^1 ((ln(x+(√(1+x^2 ))))/(√(1+x^2 )))dx solution y=ln(x+(√(1+x^2 ))) and dy=(1/(√(1+x^2 )))dx at x=0,y=0 and at x=1,y=ln(1+(√2)) hence I=∫_0 ^(ln(1+(√2))) (y/(√(1+x^2 )))×(√(1+x^2 ))dy=∫_0 ^(ln(1+(√2))) ydy I=[(y^2 /2)]_0 ^(ln(1+(√2))) =(1/2)ln^2 (1+(√2)) ∵∫_0 ^1 ((ln(x+(√(1+x^2 ))))/(√(1+x^2 )))dx=(1/2)ln^2 (1+(√2))](Q110619.png)
$${someone}\:{posted}\:{this}\:{problem}\:{and}\:{my} \\ $$$${solution}\:{followed} \\ $$$${solve} \\ $$$$\int_{\mathrm{0}} ^{\mathrm{1}} \frac{\mathrm{ln}\left({x}+\sqrt{\mathrm{1}+{x}^{\mathrm{2}} }\right)}{\sqrt{\mathrm{1}+{x}^{\mathrm{2}} }}{dx} \\ $$$${solution} \\ $$$${y}=\mathrm{ln}\left({x}+\sqrt{\mathrm{1}+{x}^{\mathrm{2}} }\right)\:\:\:{and}\:\:{dy}=\frac{\mathrm{1}}{\sqrt{\mathrm{1}+{x}^{\mathrm{2}} }}{dx} \\ $$$${at}\:{x}=\mathrm{0},{y}=\mathrm{0}\:\:{and}\:\:{at}\:\:{x}=\mathrm{1},{y}=\mathrm{ln}\left(\mathrm{1}+\sqrt{\mathrm{2}}\right) \\ $$$${hence} \\ $$$${I}=\int_{\mathrm{0}} ^{\mathrm{ln}\left(\mathrm{1}+\sqrt{\mathrm{2}}\right)} \frac{{y}}{\sqrt{\mathrm{1}+{x}^{\mathrm{2}} }}×\sqrt{\mathrm{1}+{x}^{\mathrm{2}} }{dy}=\int_{\mathrm{0}} ^{\mathrm{ln}\left(\mathrm{1}+\sqrt{\mathrm{2}}\right)} {ydy} \\ $$$${I}=\left[\frac{{y}^{\mathrm{2}} }{\mathrm{2}}\right]_{\mathrm{0}} ^{\mathrm{ln}\left(\mathrm{1}+\sqrt{\mathrm{2}}\right)} =\frac{\mathrm{1}}{\mathrm{2}}\mathrm{ln}^{\mathrm{2}} \left(\mathrm{1}+\sqrt{\mathrm{2}}\right) \\ $$$$\because\int_{\mathrm{0}} ^{\mathrm{1}} \frac{\mathrm{ln}\left({x}+\sqrt{\mathrm{1}+{x}^{\mathrm{2}} }\right)}{\sqrt{\mathrm{1}+{x}^{\mathrm{2}} }}{dx}=\frac{\mathrm{1}}{\mathrm{2}}\mathrm{ln}^{\mathrm{2}} \left(\mathrm{1}+\sqrt{\mathrm{2}}\right) \\ $$
Commented by mathdave last updated on 29/Aug/20
$${why}\:{did}\:{you}\:{mark}\:{it}\:{red}\:\:.{what}\:{is}\:{that} \\ $$$${suppose}\:\:{to}\:{mean} \\ $$
Commented by Her_Majesty last updated on 30/Aug/20
$${I}\:{didn}'{t}\:{mark}\:{it}\:{red}\:{but}\:{I}\:{can}\:{tell}\:{you}\:{the} \\ $$$${reason}\:{someone}\:{did}: \\ $$$${we}\:{have}\:\mathrm{3}\:{options}\:{to}\:{post} \\ $$$$\left(\mathrm{1}\right)\:{questions} \\ $$$$\left(\mathrm{2}\right)\:{answers} \\ $$$$\left(\mathrm{3}\right)\:{comments} \\ $$$$ \\ $$$$\left(\mathrm{1}\right)\:{is}\:{to}\:{post}\:{questions} \\ $$$$\left(\mathrm{2}\right)\:{is}\:{to}\:{post}\:{answers} \\ $$$$\left(\mathrm{3}\right)\:{is}\:{to}\:{post}\:{comments} \\ $$$$ \\ $$$${do}\:{not}\:{post}\:{questions}\:{in}\:\left(\mathrm{2}\right)\:{or}\:\left(\mathrm{3}\right) \\ $$$${do}\:{not}\:{post}\:{answers}\:{in}\:\left(\mathrm{1}\right)\:{or}\:\left(\mathrm{3}\right) \\ $$$${do}\:{not}\:{post}\:{comments}\:{in}\:\left(\mathrm{1}\right)\:{or}\:\left(\mathrm{2}\right) \\ $$$$ \\ $$$${it}'{s}\:{easy},\:{isn}'{t}\:{it}? \\ $$