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\(\int x {({x^2} + 9)^5}dx = \frac{1}{2}\int {2x} {({x^2} + 9)^5}dx = \frac{1}{2}\int {{{({x^2} + 9)}^5}d({x^2} + 9)}  = \frac{1}{2} \cdot \frac{1}{6}{({x^2} + 9)^6} + C\)