Question Number 135305 by JulioCesar last updated on 12/Mar/21
Answered by MJS_new last updated on 12/Mar/21
![∫(dx/(a^x +b))= [t=a^x +b → dx=(dt/(a^x ln a))] =(1/(ln a))∫(dt/(t(t−b)))=(1/(bln a))∫((1/(t−b))−(1/t))dt= =(1/(bln a))(ln (t−b) −ln t) = =(1/(bln a))(ln a^x −ln (a^x +b))= =(x/b)−((ln (a^x +b))/(bln a))+C](https://www.tinkutara.com/question/Q135311.png)
$$\int\frac{{dx}}{{a}^{{x}} +{b}}= \\ $$$$\:\:\:\:\:\left[{t}={a}^{{x}} +{b}\:\rightarrow\:{dx}=\frac{{dt}}{{a}^{{x}} \mathrm{ln}\:{a}}\right] \\ $$$$=\frac{\mathrm{1}}{\mathrm{ln}\:{a}}\int\frac{{dt}}{{t}\left({t}−{b}\right)}=\frac{\mathrm{1}}{{b}\mathrm{ln}\:{a}}\int\left(\frac{\mathrm{1}}{{t}−{b}}−\frac{\mathrm{1}}{{t}}\right){dt}= \\ $$$$=\frac{\mathrm{1}}{{b}\mathrm{ln}\:{a}}\left(\mathrm{ln}\:\left({t}−{b}\right)\:−\mathrm{ln}\:{t}\right)\:= \\ $$$$=\frac{\mathrm{1}}{{b}\mathrm{ln}\:{a}}\left(\mathrm{ln}\:{a}^{{x}} \:−\mathrm{ln}\:\left({a}^{{x}} +{b}\right)\right)= \\ $$$$=\frac{{x}}{{b}}−\frac{\mathrm{ln}\:\left({a}^{{x}} +{b}\right)}{{b}\mathrm{ln}\:{a}}+{C} \\ $$