Find the indefinite integral
$$\int (a+x)^{2 \over 3} x^3 dx$$
any hints?
Notice, let $a+x=t\implies dx=dt$ $$\int (a+x)^{2/3}x^3\ dx=\int t^{2/3}(t-a)^3\ dt$$ $$=\int t^{2/3}(t^3-3at^2+3a^2t-a^3)\ dt$$ $$=\int (t^{11/3}-3at^{8/3}+3a^2t^{5/3}-a^3t^{2/3})\ dt$$ $$=\int t^{11/3}\ dt-3a\int t^{8/3}\ dt+3a^2\int t^{5/3}\ dt-a^3\int t^{2/3}\ dt$$
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Notice, let $a+x=t\implies dx=dt$ $$\int (a+x)^{2/3}x^3\ dx=\int t^{2/3}(t-a)^3\ dt$$ $$=\int t^{2/3}(t^3-3at^2+3a^2t-a^3)\ dt$$ $$=\int (t^{11/3}-3at^{8/3}+3a^2t^{5/3}-a^3t^{2/3})\ dt$$ $$=\int t^{11/3}\ dt-3a\int t^{8/3}\ dt+3a^2\int t^{5/3}\ dt-a^3\int t^{2/3}\ dt$$