Question Number 7792 by Tawakalitu. last updated on 15/Sep/16
Commented by FilupSmith last updated on 16/Sep/16
![(3) ∫_1 ^( 3) x(√(x^2 −1))dx =(1/2)∫_(x=1) ^( x=3) x(x^2 −1)^(1/2) dx u=x^2 −1 du=2xdx ∫_1 ^( 3) x(√(x^2 −1))dx=(1/2)∫_(x=1) ^( x=3) (√u)du =(1/2)∫_(x=1) ^( x=3) (u)^(1/2) du =(1/2)[(u)^(3/2) (2/3)]_(x=1) ^(x=3) =(1/3)[(x^2 −1)^(3/2) ]_(x=1) ^(x=3) =(1/3)[(9−1)^(3/2) −(1−1)^(3/2) ] =(1/3)[(9−1)^(3/2) −(1−1)^(3/2) ] =(1/3)(^( 2) (√8^3 ))=(1/3)×16(√2) =((16)/3)(√2) −−−−−−−−−−−−−−−−−−−−− (4) ∫ ((e^x −e^(−x) )/(e^x +e^(−x) )) e^x +e^(−x) =u du=e^x −e^x dx =∫(1/u)du =ln(e^x +e^(−x) )+c −−−−−−−−−−−−−−−−−−−−− (5) ∫_0 ^( 3) (1/(2x+1))dx=∫_0 ^( 3) (1/2)×(2/(2x+1))dx=(1/2)∫_0 ^( 3) (2/(2x+1))dx =(1/2)[ln(2x+1)]_0 ^3 =(1/2){ln(7)−ln(1)} =(1/2)ln(7) or ∫_0 ^( 3) (1/(2x+1))dx u=2x+1 du=2dx ⇒ dx=(1/2)du ∴∫_0 ^( 3) (1/(2x+1))dx=∫_0 ^( 3) (1/u)((1/2))du same as above](https://www.tinkutara.com/question/Q7795.png)
$$\left(\mathrm{3}\right)\:\:\int_{\mathrm{1}} ^{\:\mathrm{3}} {x}\sqrt{{x}^{\mathrm{2}} −\mathrm{1}}{dx} \\ $$$$=\frac{\mathrm{1}}{\mathrm{2}}\int_{{x}=\mathrm{1}} ^{\:{x}=\mathrm{3}} {x}\left({x}^{\mathrm{2}} −\mathrm{1}\right)^{\frac{\mathrm{1}}{\mathrm{2}}} {dx} \\ $$$${u}={x}^{\mathrm{2}} −\mathrm{1} \\ $$$${du}=\mathrm{2}{xdx} \\ $$$$\int_{\mathrm{1}} ^{\:\mathrm{3}} {x}\sqrt{{x}^{\mathrm{2}} −\mathrm{1}}{dx}=\frac{\mathrm{1}}{\mathrm{2}}\int_{{x}=\mathrm{1}} ^{\:{x}=\mathrm{3}} \sqrt{{u}}{du} \\ $$$$=\frac{\mathrm{1}}{\mathrm{2}}\int_{{x}=\mathrm{1}} ^{\:{x}=\mathrm{3}} \left({u}\right)^{\frac{\mathrm{1}}{\mathrm{2}}} {du} \\ $$$$=\frac{\mathrm{1}}{\mathrm{2}}\left[\left({u}\right)^{\frac{\mathrm{3}}{\mathrm{2}}} \frac{\mathrm{2}}{\mathrm{3}}\right]_{{x}=\mathrm{1}} ^{{x}=\mathrm{3}} \\ $$$$=\frac{\mathrm{1}}{\mathrm{3}}\left[\left({x}^{\mathrm{2}} −\mathrm{1}\right)^{\frac{\mathrm{3}}{\mathrm{2}}} \right]_{{x}=\mathrm{1}} ^{{x}=\mathrm{3}} \\ $$$$=\frac{\mathrm{1}}{\mathrm{3}}\left[\left(\mathrm{9}−\mathrm{1}\right)^{\frac{\mathrm{3}}{\mathrm{2}}} −\left(\mathrm{1}−\mathrm{1}\right)^{\frac{\mathrm{3}}{\mathrm{2}}} \right] \\ $$$$=\frac{\mathrm{1}}{\mathrm{3}}\left[\left(\mathrm{9}−\mathrm{1}\right)^{\frac{\mathrm{3}}{\mathrm{2}}} −\left(\mathrm{1}−\mathrm{1}\right)^{\frac{\mathrm{3}}{\mathrm{2}}} \right] \\ $$$$=\frac{\mathrm{1}}{\mathrm{3}}\left(^{\:\mathrm{2}} \sqrt{\mathrm{8}^{\mathrm{3}} }\right)=\frac{\mathrm{1}}{\mathrm{3}}×\mathrm{16}\sqrt{\mathrm{2}} \\ $$$$=\frac{\mathrm{16}}{\mathrm{3}}\sqrt{\mathrm{2}} \\ $$$$ \\ $$$$−−−−−−−−−−−−−−−−−−−−− \\ $$$$ \\ $$$$\left(\mathrm{4}\right)\:\int\:\frac{{e}^{{x}} −{e}^{−{x}} }{{e}^{{x}} +{e}^{−{x}} } \\ $$$${e}^{{x}} +{e}^{−{x}} ={u} \\ $$$${du}={e}^{{x}} −{e}^{{x}} \:{dx} \\ $$$$=\int\frac{\mathrm{1}}{{u}}{du} \\ $$$$=\mathrm{ln}\left({e}^{{x}} +{e}^{−{x}} \right)+{c} \\ $$$$ \\ $$$$−−−−−−−−−−−−−−−−−−−−− \\ $$$$ \\ $$$$\left(\mathrm{5}\right)\:\int_{\mathrm{0}} ^{\:\mathrm{3}} \frac{\mathrm{1}}{\mathrm{2}{x}+\mathrm{1}}{dx}=\int_{\mathrm{0}} ^{\:\mathrm{3}} \frac{\mathrm{1}}{\mathrm{2}}×\frac{\mathrm{2}}{\mathrm{2}{x}+\mathrm{1}}{dx}=\frac{\mathrm{1}}{\mathrm{2}}\int_{\mathrm{0}} ^{\:\mathrm{3}} \frac{\mathrm{2}}{\mathrm{2}{x}+\mathrm{1}}{dx} \\ $$$$=\frac{\mathrm{1}}{\mathrm{2}}\left[\mathrm{ln}\left(\mathrm{2}{x}+\mathrm{1}\right)\right]_{\mathrm{0}} ^{\mathrm{3}} \\ $$$$=\frac{\mathrm{1}}{\mathrm{2}}\left\{\mathrm{ln}\left(\mathrm{7}\right)−\mathrm{ln}\left(\mathrm{1}\right)\right\} \\ $$$$=\frac{\mathrm{1}}{\mathrm{2}}\mathrm{ln}\left(\mathrm{7}\right) \\ $$$$ \\ $$$${or} \\ $$$$ \\ $$$$\int_{\mathrm{0}} ^{\:\mathrm{3}} \frac{\mathrm{1}}{\mathrm{2}{x}+\mathrm{1}}{dx} \\ $$$${u}=\mathrm{2}{x}+\mathrm{1} \\ $$$${du}=\mathrm{2}{dx}\:\:\Rightarrow\:\:{dx}=\frac{\mathrm{1}}{\mathrm{2}}{du} \\ $$$$\therefore\int_{\mathrm{0}} ^{\:\mathrm{3}} \frac{\mathrm{1}}{\mathrm{2}{x}+\mathrm{1}}{dx}=\int_{\mathrm{0}} ^{\:\mathrm{3}} \frac{\mathrm{1}}{{u}}\left(\frac{\mathrm{1}}{\mathrm{2}}\right){du} \\ $$$${same}\:{as}\:{above} \\ $$
Commented by Tawakalitu. last updated on 16/Sep/16
$${Thank}\:{you}\:{sir}.\:{for}\:{your}\:{help}. \\ $$
Commented by Tawakalitu. last updated on 16/Sep/16
$${Thanks}\:{so}\:{much}\:{sir},\:{i}\:{really}\:{appreciate}. \\ $$