Indefinite integrals: Method of supstitution

\[\int{x{{\left( 5{{x}^{2}}-3 \right)}^{7}}dx=\left| \begin{align} & t=5{{x}^{2}}-3/' \\ & dt=10xdx/:10 \\ & xdx=\frac{dt}{10} \\ \end{align} \right|=}\frac{1}{10}\int{{{t}^{7}}dt=\frac{1}{10}\frac{{{t}^{8}}}{8}+C=\frac{{{\left( 5{{x}^{2}}-3 \right)}^{8}}}{80}}+C\] \[\int{\frac{xdx}{\sqrt{x+1}}}=\left| \begin{align} & t=\sqrt{x+1}/' \\ & dt=\frac{dx}{2\sqrt{x+1}} \\ & 2dt=\frac{dx}{\sqrt{x+1}} \\ & x={{t}^{2}}-1 \\ \end{align} \right|=\int{2\left( {{t}^{2}}-1 \right)}dt=2\int{{{t}^{2}}dt}-2\int{dt}=\frac{2}{3}{{t}^{3}}-2t+C=\frac{2}{3}{{\left( \sqrt{x+1} \right)}^{3}}-2\sqrt{x+1}+C\] \[\int{\frac{\cos xdx}{\sqrt{1+{{\sin }^{2}}x}}=\left| \begin{align} & t=\sin x/' \\ & dt=\cos xdx \\ \end{align} \right|=\int{\frac{dt}{\sqrt{1+{{t}^{2}}}}}}=\ln \left| t+\sqrt{{{t}^{2}}+1} \right|+C=\ln \left| \sin x+\sqrt{{{\sin }^{2}}x+1} \right|+C\] \[\int{x{{\left( 2x+5 \right)}^{10}}dx}=\left| \begin{align} & t=2x+5/' \\ & dt=2dx \\ & dx=\frac{1}{2}dt \\ & x=\frac{1}{2}\left( t-5 \right) \\ \end{align} \right|=\frac{1}{4}\int{\left( {{t}^{11}}-5{{t}^{10}} \right)dt=\frac{1}{4}\frac{{{t}^{12}}}{12}}-\frac{5}{11}{{t}^{11}}+C=\frac{{{\left( 2x+5 \right)}^{12}}}{48}-\frac{5{{\left( 2x+5 \right)}^{11}}}{11}+C\]