1Integration by parts

1 ln x dx = Z x0ln x dx = x ln x Z x 1 x dx = x ln x x +C. Example 1.4 09 September Evaluate R ex sin x dx. Solution Neither factor simplifies much when differentiated, but ex is easier to integrate, so let’s try passing the prime from it onto sin x: Z ex sin x dx = Z (ex)0sin x dx = ex sin x Z ................
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