Indefinite Integral DOUBT Questions

\hspace{-16}\bf{(2)::\int \frac{1}{(x+3)^{\frac{8}{7}}.(x-2)^{\frac{6}{7}}}dx}$\\\\\\ $\bf{\int \frac{1}{\big(\frac{x+3}{x-2}\big)^{\frac{8}{7}}}.\frac{1}{(x-2)^2}dx}$\\\\\\ Now Put $\bf{\frac{x+3}{x-2}=t\Leftrightarrow \frac{-5}{(x-2)^2}dx = dt}$\\\\\\ Or $\bf{\frac{1}{(x-2)^2}dx = -\frac{1}{5}dt}$\\\\\\ So $\bf{-\frac{1}{5}.\int \frac{1}{t^{\frac{8}{7}}}dt=-\frac{1}{5}.\frac{-7}{1}.t^{\frac{-1}{7}}+C}$\\\\\\ $\bf{=\frac{7}{5}.\left(\frac{x-2}{x+3}\right)^{\frac{1}{7}}+C}$

\hspace{-16}\bf{(1)\;\;\int \frac{1}{x^{22}.(x^7-6)}dx}$\\\\\\ $\bf{\int\frac{1}{x^{29}.(1-6x^{-7})}dx}$\\\\\\ Now Sub. $\bf{6x^{-7}=t\Leftrightarrow -42x^{-8}dx=dt}$\\\\\\ $\bf{dx=-\frac{1}{42}.x^{8}dt}$\\\\\\ So $\bf{-\frac{1}{42}.\int\frac{1}{x^{21}.(1-t)}dt=-\frac{1}{42}.\int\frac{t^3}{216.(1-t)}dt}$\\\\\\ $\bf{\frac{1}{7.6^4}\int\frac{t^3}{t-1}dt=\frac{1}{7.6^4}\int \frac{(t^3-1)+1}{t-1}dt}$\\\\\\ $\bf{\frac{1}{7.6^4}\int t^2+t+1+\frac{1}{7.6^4}.\int\frac{1}{t-1}dt}$\\\\\\ $\bf{\frac{1}{7.6^4}\left(t^3+t^2+t\right)+\frac{1}{7.6^4}\ln\mid t-1\mid+C}$\\\\\\ $\bf{=\frac{1}{7.6^4}\left\{(6x^{-7})^3+(6x^{-7})^2+(6x^{-7})\right\}+\frac{1}{7.6^4}\ln \mid (6x^{-7})-1\mid+C}$

For 3rd, we get ∫ dx/(ln x +1) , whose integration is currently above my mind.

\hspace{-16}\bf{(1)\;\; \int \frac{\ln(x)-1}{\left(\ln(x)-1\right)^2}dx}$\\\\\\ $\bf{\int \frac{1}{\ln(x)+1}dx=\int \frac{1}{\ln(e.x)}dx}$\\\\\\ Put $\bf{\ln(ex)=t\Leftrightarrow ex=e^t\Leftrightarrow edx=e^tdt}$\\\\\\ $\bf{\frac{1}{e}.\int \frac{e^t}{t}dt=\frac{1}{e}\int \frac{1+\frac{t}{1!}+\frac{t^2}{2!}+......\infty}{t}dt}$\\\\\\ $\bf{\frac{1}{e}.\left\{\ln(t)+\frac{t}{1!}+\frac{t^2}{2.2!}...........\right\}+C}$\\\\\\ $\bf{\frac{1}{e}.\left\{\ln(\ln(ex))+\frac{\ln(ex)}{1!}+\frac{(\ln(ex))^2}{2.2!}...........\right\}+C}$