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If u 8+x is substituted into x3(8+)6 dx, then we have (x (1) dx = [ 10 x dx. We must also convert x3
If u 8+x is substituted into x3(8+)6 dx, then we have (x (1) dx = [ 10 x dx. We must also convert x3 dx into an expression involving u. We know that du = 4x dx, and so x3 dx = du. (u)6
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