Question-226178

Question Number 226178 by Spillover last updated on 22/Nov/25

Answered by mr W last updated on 22/Nov/25

=∫((1+(1/x))/(x+(1/x)))dx =∫((x+1)/(x^2 +1))dx =(1/2)∫((d(x^2 +1))/(x^2 +1))+∫(1/(x^2 +1))dx =(1/2)ln (x^2 +1)+tan^(−1) x+C ∫_1 ^3 =[(1/2)ln (x^2 +1)+tan^(−1) x]_1 ^3 =(1/2)ln 5+tan^(−1) 3−(π/4)

$$=\int\frac{\mathrm{1}+\frac{\mathrm{1}}{{x}}}{{x}+\frac{\mathrm{1}}{{x}}}{dx} \\ $$$$=\int\frac{{x}+\mathrm{1}}{{x}^{\mathrm{2}} +\mathrm{1}}{dx} \\ $$$$=\frac{\mathrm{1}}{\mathrm{2}}\int\frac{{d}\left({x}^{\mathrm{2}} +\mathrm{1}\right)}{{x}^{\mathrm{2}} +\mathrm{1}}+\int\frac{\mathrm{1}}{{x}^{\mathrm{2}} +\mathrm{1}}{dx} \\ $$$$=\frac{\mathrm{1}}{\mathrm{2}}\mathrm{ln}\:\left({x}^{\mathrm{2}} +\mathrm{1}\right)+\mathrm{tan}^{−\mathrm{1}} {x}+{C} \\ $$$$\int_{\mathrm{1}} ^{\mathrm{3}} =\left[\frac{\mathrm{1}}{\mathrm{2}}\mathrm{ln}\:\left({x}^{\mathrm{2}} +\mathrm{1}\right)+\mathrm{tan}^{−\mathrm{1}} {x}\right]_{\mathrm{1}} ^{\mathrm{3}} \\ $$$$\:=\frac{\mathrm{1}}{\mathrm{2}}\mathrm{ln}\:\mathrm{5}+\mathrm{tan}^{−\mathrm{1}} \mathrm{3}−\frac{\pi}{\mathrm{4}} \\ $$

Commented by fantastic2 last updated on 22/Nov/25