Question Number 49225 by AdqhK ÐQeQqQ last updated on 04/Dec/18
$$\int\frac{\mathrm{1}}{{z}\left({z}^{\mathrm{17}} +\mathrm{1}\right)}=??\:{find}\:{please} \\ $$
Answered by arvinddayama01@gmail.com. last updated on 04/Dec/18
![∫(1/(z(z^(17) +1))).(z^(16) /z^(16) )dz z^(17) =t d.w.r.to z z^(16) dz=(1/(17))dt (1/(17))∫(dt/(t(t+1))) (1/(17))∫[(1/t)−(1/(t+1))]dt (1/(17))ln((t/(t+1)))+C (1/(17))ln((z^(17) /(z^(17) +1)))+C ∵∫(1/(x(x^n +1)))dx=(1/n)ln((x^n /(x^n +1)))+C](https://www.tinkutara.com/question/Q49226.png)
$$\int\frac{\mathrm{1}}{{z}\left({z}^{\mathrm{17}} +\mathrm{1}\right)}.\frac{{z}^{\mathrm{16}} }{{z}^{\mathrm{16}} }{dz} \\ $$$${z}^{\mathrm{17}} ={t} \\ $$$${d}.{w}.{r}.{to}\:{z} \\ $$$${z}^{\mathrm{16}} {dz}=\frac{\mathrm{1}}{\mathrm{17}}{dt} \\ $$$$\frac{\mathrm{1}}{\mathrm{17}}\int\frac{{dt}}{{t}\left({t}+\mathrm{1}\right)} \\ $$$$\frac{\mathrm{1}}{\mathrm{17}}\int\left[\frac{\mathrm{1}}{{t}}−\frac{\mathrm{1}}{{t}+\mathrm{1}}\right]{dt} \\ $$$$\frac{\mathrm{1}}{\mathrm{17}}{ln}\left(\frac{{t}}{{t}+\mathrm{1}}\right)+{C} \\ $$$$\frac{\mathrm{1}}{\mathrm{17}}{ln}\left(\frac{{z}^{\mathrm{17}} }{{z}^{\mathrm{17}} +\mathrm{1}}\right)+{C} \\ $$$$\because\int\frac{\mathrm{1}}{{x}\left({x}^{{n}} +\mathrm{1}\right)}{dx}=\frac{\mathrm{1}}{{n}}{ln}\left(\frac{{x}^{{n}} }{{x}^{{n}} +\mathrm{1}}\right)+{C} \\ $$
Commented by AdqhK ÐQeQqQ last updated on 04/Dec/18
$${thanku}\:{sir} \\ $$$$ \\ $$