MATH 110 Lecture 47: Examples

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16 Nov 2015
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Evaluate dx u x dx d = 2. 1 eu du u + c eu x x2 e. Evaluate x x2 + ( x e e /(e x + 3. U = x2 u x dx d = 2 du u + c eu. = ( 1 x + 3 1 ex dx. ) dx (e u = ex + 3 (1/(e (1/u) du n l u| /2)e d = ex u dx x + 3 x. | + c = ln e| x + 3| + c. X2 + l e| x + 3| + c n to complete the substitution: t = u 1. /(t t + 1 u = t + 1. Solve for (u u l u| n. = x dx x2 = u 1 t| + 1| + c = t l n t| + 1| + c x3 u = x2 + 1 (u. 2 + 1 5/2 + c u5/2 u3/2 du.

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