Question Number 226702 by leromain last updated on 10/Dec/25
$$\int\frac{{ln}\left({x}^{\mathrm{2}} +\mathrm{3}{x}+\mathrm{2}\right)}{{x}^{\mathrm{2}} +\mathrm{1}}{dx}\left[\right. \\ $$
Answered by Frix last updated on 11/Dec/25
![∫((ln (x^2 +3x+2))/(x^2 +1))dx=∫((ln ((x+1)(x+2)))/(x^2 +1))dx= =∫((ln (x+1))/(x^2 +1))dx+∫((ln (x+2))/(x^2 +1))dx Here′s the path, complete it by inserting: 1. ∫((ln (x+a))/(x^2 +1))dx=∫((ln (x+a))/((x−i)(x+i)))dx= =(i/2)∫(((ln (x+a))/(x+i))−((ln (x+a))/(x−i)))dx 2. ∫((ln (x+a))/(x+b))dx =^([t=x+b]) =∫((ln (t+a−b))/t)dt=∫((ln ((((a−b)t)/(a−b))+(((a−b)^2 )/(a−b))))/t)dt= =∫(((ln (a−b))/t)+((ln ((t/(a−b))+1))/t))dt 2.1. ∫((ln (a−b))/t)dt=ln (a−b) ln t 2.2. ∫((ln ((t/(a−b))+1))/t)dt =^([u=−(t/(a−b))]) ∫((ln (1−u))/u)du= =−Li_2 (u) =−Li_2 ((t/(b−a)))](Q226712.png)
$$\int\frac{\mathrm{ln}\:\left({x}^{\mathrm{2}} +\mathrm{3}{x}+\mathrm{2}\right)}{{x}^{\mathrm{2}} +\mathrm{1}}{dx}=\int\frac{\mathrm{ln}\:\left(\left({x}+\mathrm{1}\right)\left({x}+\mathrm{2}\right)\right)}{{x}^{\mathrm{2}} +\mathrm{1}}{dx}= \\ $$$$=\int\frac{\mathrm{ln}\:\left({x}+\mathrm{1}\right)}{{x}^{\mathrm{2}} +\mathrm{1}}{dx}+\int\frac{\mathrm{ln}\:\left({x}+\mathrm{2}\right)}{{x}^{\mathrm{2}} +\mathrm{1}}{dx} \\ $$$$ \\ $$$$\mathrm{Here}'\mathrm{s}\:\mathrm{the}\:\mathrm{path},\:\mathrm{complete}\:\mathrm{it}\:\mathrm{by}\:\mathrm{inserting}: \\ $$$$\mathrm{1}. \\ $$$$\int\frac{\mathrm{ln}\:\left({x}+{a}\right)}{{x}^{\mathrm{2}} +\mathrm{1}}{dx}=\int\frac{\mathrm{ln}\:\left({x}+{a}\right)}{\left({x}−\mathrm{i}\right)\left({x}+\mathrm{i}\right)}{dx}= \\ $$$$=\frac{\mathrm{i}}{\mathrm{2}}\int\left(\frac{\mathrm{ln}\:\left({x}+{a}\right)}{{x}+\mathrm{i}}−\frac{\mathrm{ln}\:\left({x}+{a}\right)}{{x}−\mathrm{i}}\right){dx} \\ $$$$ \\ $$$$\mathrm{2}. \\ $$$$\int\frac{\mathrm{ln}\:\left({x}+{a}\right)}{{x}+{b}}{dx}\:\overset{\left[{t}={x}+{b}\right]} {=} \\ $$$$=\int\frac{\mathrm{ln}\:\left({t}+{a}−{b}\right)}{{t}}{dt}=\int\frac{\mathrm{ln}\:\left(\frac{\left({a}−{b}\right){t}}{{a}−{b}}+\frac{\left({a}−{b}\right)^{\mathrm{2}} }{{a}−{b}}\right)}{{t}}{dt}= \\ $$$$=\int\left(\frac{\mathrm{ln}\:\left({a}−{b}\right)}{{t}}+\frac{\mathrm{ln}\:\left(\frac{{t}}{{a}−{b}}+\mathrm{1}\right)}{{t}}\right){dt} \\ $$$$ \\ $$$$\mathrm{2}.\mathrm{1}. \\ $$$$\int\frac{\mathrm{ln}\:\left({a}−{b}\right)}{{t}}{dt}=\mathrm{ln}\:\left({a}−{b}\right)\:\mathrm{ln}\:{t} \\ $$$$ \\ $$$$\mathrm{2}.\mathrm{2}. \\ $$$$\int\frac{\mathrm{ln}\:\left(\frac{{t}}{{a}−{b}}+\mathrm{1}\right)}{{t}}{dt}\:\overset{\left[{u}=−\frac{{t}}{{a}−{b}}\right]} {=}\:\int\frac{\mathrm{ln}\:\left(\mathrm{1}−{u}\right)}{{u}}{du}= \\ $$$$=−\mathrm{Li}_{\mathrm{2}} \:\left({u}\right)\:=−\mathrm{Li}_{\mathrm{2}} \:\left(\frac{{t}}{{b}−{a}}\right) \\ $$