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Question Number 141530 by sarkor last updated on 20/May/21

Commented by mohammad17 last updated on 20/May/21

$$=\int \frac{x^2-9+9}{\sqrt{x-3}}dx=\int \frac{(x-3)(x+3)}{\sqrt{x-3}}dx+\int \frac{9}{\sqrt{x-3}}dx$$$$$$$$=\int \left(\sqrt{x-3}\right)(x+3)dx+\int 9{(x-3)}^{-\frac{1}{2}}dx$$$$$$$$=\int \left(\sqrt{x-3}\right)(x-3+6)dx+9\int {(x-3)}^{-\frac{1}{2}}dx$$$$$$$$=\int {(x-3)}^{\frac{3}{2}}dx+6\int {(x-3)}^{\frac{1}{2}}dx+9\int {(x-3)}^{-\frac{1}{2}}dx$$$$$$$$=\frac{2}{5}\sqrt{{(x-3)}^5}+4\sqrt{{(x-3)}^3}+18\sqrt{(x-3)}+C$$$$$$$$by::⟨m.o⟩$$

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