Indefinite Integrals

\[\int{5{{a}^{2}}{{x}^{6}}dx=5{{a}^{2}}\int{{{x}^{6}}dx}}=5{{a}^{2}}\frac{{{x}^{6+1}}}{6+1}+C=\frac{{{x}^{7}}}{7}+C=\frac{5}{7}{{a}^{2}}{{x}^{7}}+C\] \[\begin{align} & \int{\left( 6{{x}^{2}}+8x+3 \right)dx=\int{6{{x}^{2}}dx+}}\int{8xdx+}\int{3dx}=6\int{{{x}^{2}}dx}+8\int{xdx}+3\int{{{x}^{0}}dx}= \\ & 6\frac{{{x}^{2+1}}}{2+1}+8\frac{{{x}^{1+1}}}{1+1}+3\frac{{{x}^{0+1}}}{0+1}+C=6\frac{{{x}^{3}}}{3}+8\frac{{{x}^{2}}}{2}+3x+C=2{{x}^{3}}+4{{x}^{2}}+3x+C \\ \end{align}\] \[\begin{align} & \int{x\left( x+a \right)\left( x+b \right)}dx=\int{x\left( {{x}^{2}}+ax+ax+ab \right)}dx=\int{\left( {{x}^{3}}+2a{{x}^{2}}+abx \right)}dx= \\ & =\int{{{x}^{3}}}dx+2a\int{{{x}^{2}}}dx+ab\int{x}dx=\frac{{{x}^{4}}}{4}+2a\frac{{{x}^{3}}}{3}+ab\frac{x}{2}+C=\frac{{{x}^{4}}}{4}+\frac{2}{3}a{{x}^{3}}+\frac{1}{2}abx+C \\ \end{align}\] \[\begin{align} & \int{{{\left( a+b{{x}^{3}} \right)}^{2}}dx=}\int{\left( {{a}^{2}}+2ab{{x}^{3}}+{{b}^{2}}{{x}^{6}} \right)}dx={{a}^{2}}\int{dx}+2ab\int{{{x}^{3}}dx}+{{b}^{2}}\int{{{x}^{6}}dx}= \\ & ={{a}^{2}}x+2ab\frac{{{x}^{4}}}{4}+{{b}^{2}}\frac{{{x}^{7}}}{7}+C={{a}^{2}}x+\frac{1}{2}ab{{x}^{4}}+\frac{1}{7}{{b}^{2}}{{x}^{7}}+C \\ \end{align}\] \[\int{\sqrt{2px}}dx=\sqrt{2p}\int{\sqrt{x}dx}=\sqrt{2p}\int{{{x}^{\frac{1}{2}}}dx}=\sqrt{2p}\frac{{{x}^{\frac{1}{2}+1}}}{\frac{1}{2}+1}+C=\sqrt{2p}\frac{{{x}^{\frac{3}{2}}}}{\frac{3}{2}}+C=\frac{2\sqrt{2p}}{3}\sqrt{{{x}^{3}}}+C\] \[\int{\frac{dx}{\sqrt[n]{x}}}=\int{{{x}^{-\frac{1}{n}}}}dx=\frac{{{x}^{\frac{-1+n}{n}}}}{\frac{-1+n}{n}}+C=\frac{n}{n-1}\sqrt[n]{{{x}^{n-1}}}+C\] \[\begin{align} & \int{{{\left( nx \right)}^{\frac{1-n}{n}}}dx}={{n}^{\frac{1-n}{n}}}\int{{{x}^{\frac{1-n}{n}}}dx}={{n}^{\frac{1-n}{n}}}\frac{{{x}^{\frac{1-n}{n}+1}}}{\frac{1-n}{n}+1}+C={{n}^{\frac{1-n}{n}}}\frac{{{x}^{\frac{1-n+n}{n}}}}{\frac{1-n+n}{n}}+C={{n}^{\frac{1-n}{n}}}\frac{{{x}^{\frac{1}{n}}}}{\frac{1}{n}}+C= \\ & ={{n}^{\frac{1-n}{n}}}{{n}^{1}}\sqrt[n]{x}+C={{n}^{\frac{1-n+n}{n}}}\sqrt[n]{x}+C=\sqrt[n]{nx}+C \\ \end{align}\] \[\begin{align} & {{\int{\left( {{a}^{\frac{2}{3}}}-{{x}^{\frac{2}{3}}} \right)}}^{3}}dx=\int{\left( {{a}^{2}}-3{{a}^{{4}/{3}\;}}{{x}^{{2}/{3}\;}}+3{{a}^{{2}/{3}\;}}{{x}^{{4}/{3}\;}}-{{x}^{2}} \right)}dx={{a}^{2}}\int{dx}-3{{a}^{{4}/{3}\;}}\int{{{x}^{{2}/{3}\;}}dx}+3{{a}^{{2}/{3}\;}}\int{{{x}^{{4}/{3}\;}}dx}-\int{{{x}^{2}}dx}= \\ & ={{a}^{2}}x-3{{a}^{{4}/{3}\;}}\frac{{{x}^{\frac{2}{3}+1}}}{\frac{2}{3}+1}+3{{a}^{{2}/{3}\;}}\frac{{{x}^{\frac{4}{3}+1}}}{\frac{4}{3}+1}-\frac{{{x}^{3}}}{3}+C={{a}^{2}}x-3{{a}^{{4}/{3}\;}}\frac{{{x}^{\frac{5}{3}}}}{\frac{5}{3}}+3{{a}^{{2}/{3}\;}}\frac{{{x}^{\frac{7}{3}}}}{\frac{7}{3}}-\frac{{{x}^{3}}}{3}+C= \\ & ={{a}^{2}}x-\frac{9}{5}{{a}^{{4}/{3}\;}}{{x}^{\frac{5}{3}}}+\frac{9}{7}{{a}^{{4}/{3}\;}}{{x}^{\frac{7}{3}}}-\frac{{{x}^{3}}}{3}+C \\ \end{align}\] \[\begin{align} & \int{\left( \sqrt{x}+1 \right)\left( x-\sqrt{x}+1 \right)}dx=\int{\left( x\sqrt{x}-x+\sqrt{x}+x-\sqrt{x}+1 \right)}dx=\int{\left( x\sqrt{x}+1 \right)}dx= \\ & =\int{x\sqrt{x}}dx+\int{dx}=\int{{{x}^{\frac{3}{2}}}dx}+\int{dx}=\frac{{{x}^{\frac{3}{2}+1}}}{\frac{3}{2}+1}+x+C=\frac{{{x}^{\frac{5}{2}}}}{\frac{5}{2}}+x+C=\frac{2}{5}\sqrt{{{x}^{5}}}+x+C \\ \end{align}\] \[\begin{align} & \int{\frac{\left( {{x}^{2}}+1 \right)\left( {{x}^{2}}-2 \right)}{\sqrt[3]{{{x}^{2}}}}}dx=\int{\frac{\left( {{x}^{4}}-2{{x}^{2}}+{{x}^{2}}-2 \right)}{\sqrt[3]{{{x}^{2}}}}}dx=\int{\left( {{x}^{4}}-{{x}^{2}}-2 \right)}{{x}^{-\frac{2}{3}}}dx=\int{\left( {{x}^{4-\frac{2}{3}}}-{{x}^{2-\frac{2}{3}}}-2{{x}^{-\frac{2}{3}}} \right)}dx \\ & \int{{{x}^{\frac{12-2}{3}}}}dx-\int{{{x}^{\frac{6-2}{3}}}dx}-2\int{{{x}^{-\frac{2}{3}}}}dx=\int{{{x}^{\frac{10}{3}}}dx-\int{{{x}^{\frac{4}{3}}}}}dx-2\int{{{x}^{-\frac{2}{3}}}}dx=\frac{{{x}^{\frac{10}{3}+1}}}{\frac{10}{3}+1}-\frac{{{x}^{\frac{4}{3}+1}}}{\frac{4}{3}+1}-2\frac{{{x}^{-\frac{2}{3}+1}}}{-\frac{2}{3}+1}+C \\ & =\frac{{{x}^{\frac{13}{3}}}}{\frac{13}{3}}-\frac{{{x}^{\frac{7}{3}}}}{\frac{7}{3}}-2\frac{{{x}^{\frac{1}{3}}}}{\frac{1}{3}}+C=\frac{3}{13}{{x}^{\frac{13}{3}}}-\frac{3}{7}{{x}^{\frac{7}{3}}}-6{{x}^{\frac{1}{3}}}+C \\ \end{align}\] \[\begin{align} & \int{\frac{{{\left( {{x}^{m}}-{{x}^{n}} \right)}^{2}}}{\sqrt{x}}}dx=\int{\left( {{x}^{2m}}-2{{x}^{m+n}}+{{x}^{2n}} \right)}{{x}^{-\frac{1}{2}}}dx=\int{{{x}^{2m-\frac{1}{2}}}}dx-2\int{{{x}^{m+n-\frac{1}{2}}}}dx+\int{{{x}^{2n-\frac{1}{2}}}}dx= \\ & =\frac{{{x}^{2m-\frac{1}{2}+1}}}{2m-\frac{1}{2}+1}-2\frac{{{x}^{m+n-\frac{1}{2}+1}}}{m+n-\frac{1}{2}+1}+\frac{{{x}^{2n-\frac{1}{2}+1}}}{2n-\frac{1}{2}+1}+C=\frac{{{x}^{\frac{4m-1+2}{2}}}}{\frac{4m-1+2}{2}}-2\frac{{{x}^{\frac{2m+2n-1+2}{2}}}}{\frac{2m+2n-1+2}{2}}+\frac{{{x}^{\frac{4n-1+2}{2}}}}{\frac{4n-1+2}{2}}+C= \\ & =\frac{2}{4m+1}{{x}^{\frac{4m+1}{2}}}-2\frac{2}{2m+2n+1}{{x}^{\frac{2m+2n+1}{2}}}+\frac{2}{4n+1}{{x}^{\frac{4n+1}{2}}}+C \\ \end{align}\] \[\int{\frac{dx}{{{x}^{2}}+7}=}\frac{1}{\sqrt{7}}\arctan \frac{x}{\sqrt{7}}+C\] \[\int{\frac{dx}{{{x}^{2}}-10}=}\frac{1}{2\sqrt{10}}\ln \left| \frac{x-\sqrt{10}}{x+\sqrt{10}} \right|+C\] \[\int{\frac{dx}{\sqrt{4+{{x}^{2}}}}=}\ln \left| x+\sqrt{{{x}^{2}}+4} \right|+C\] \[\int{\frac{dx}{\sqrt{8-{{x}^{2}}}}=}\arcsin \frac{x}{8}+C\] \[\begin{align} & \int{\frac{\sqrt{2+{{x}^{2}}}-\sqrt{2-{{x}^{2}}}}{\sqrt{4-{{x}^{4}}}}dx=}\int{\frac{\sqrt{2+{{x}^{2}}}}{\sqrt{\left( 2+{{x}^{2}} \right)\left( 2-{{x}^{2}} \right)}}}dx-\int{\frac{\sqrt{2-{{x}^{2}}}}{\sqrt{\left( 2+{{x}^{2}} \right)\left( 2-{{x}^{2}} \right)}}dx} \\ & =\int{\frac{dx}{\sqrt{2-{{x}^{2}}}}}-\int{\frac{dx}{\sqrt{2+{{x}^{2}}}}}=\arcsin \frac{x}{\sqrt{2}}-\ln \left| x+\sqrt{{{x}^{2}}+2} \right|+C \\ \end{align}\]