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Extra resources for Calculus 1c-3, Examples of Integrals

Example text

D. Introduce the indicated substitutions. √ I. 1) The substitution t = x, x > 0, t > 0, is monotonous, x = t2 , and dx = 2t dt. Hence √ 1 − 2t 2t − 4t2 1−2 x √ dx = · 2t dt = dt. √ √ 1+3 x t= x 1 + 3t t= x 1 + 3t A division by polynomials gives 2t − 4t2 4 10 10 10 1 =− t+ − − · , 1 + 3t 3 9 9 27 t + 13 hence by insertion, √ 1−2 x √ dx = 1+3 x √ t= x 10 10 1 4 − · − t+ 3 9 27 t + 1 3 dt 10 2 10 t− ln(3t + 1) √ − t2 + 3 9 27 t= x √ 10 √ 2 10 ln(1 + 3 x). com 32 Calculus 1c-3 Integration by simple substitutes C.

Test. D. 4 Find an integral of the function ⎧ 3 ⎨ 1 x+2 4 x + 1 f (x) = −2+ 4 (x + 2)2 ⎩ x+2 x+1 3 ⎫ ⎬ ⎭ , −1 < x < ∞. A. Integration. D. A substitution followed by an integration. I. e. x= 2u4 − 1 , 1 − u4 and 8u3 (1 − u4 ) + 4u3 (2u4 − 1) du (1 − u4 )2 2 − 2u4 + 2u4 − 1 4u3 du = du. com 41 Calculus 1c-3 Since Integration by advanced substitutions 1 = 1 − u4 , we get x+2 ⎧ 3 ⎨ 1 4 x + 1 −2+ (x + 2)2 ⎩ x+2 = u= = u= 4 = 7 q 4 x+1 x+2 (1 − u4 )2 u3 − 2 + u−3 · q 4 x+1 x+2 4 4 x+2 x+1 4u6 − 8u3 + 4 du = x+1 x+2 7 −2· 3 ⎫ ⎬ ⎭ dx 4u3 du (1 − u4 )2 4 7 u − 2u4 + 4u 7 u= q 4 x+1 x+2 x+1 x+1 +44 .

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