IntegrationReviewSolns - Integration Review MATH 213-C...

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Unformatted text preview: Integration Review MATH 213-C September 25, 2009 Evaluate the integrals. 1. Z 1 a 2 + x 2 dx,a 6 = 0 Solution: The integrand almost looks like the derivative of tan- 1 ( x ). Factor an a 2 out of the denominator: Z 1 a 2 + x 2 dx = 1 a 2 Z 1 1 + ( x a ) 2 dx Let u = x a du = 1 a dx Thus Z 1 a 2 + x 2 dx = 1 a Z 1 1 + u 2 du = 1 a tan- 1 x a + c 2. Z 1 - 5 + 6 x- x 2 dx Solution: Complete the square:- x 2 + 6 x- 5 =- ( x 2 + 6 x + 9) + 4 =- ( x- 3) 2 + 4 1 so Z 1 - 5 + 6 x- x 2 dx = Z 1 p- ( x- 3) 2 + 4 dx = 1 2 Z 1 q 1- ( x- 3 2 ) 2 dx, for x- 3 2 < 1 Let u = x- 3 2 du = 1 2 dx so that Z 1 - 5 + 6 x- x 2 dx = Z 1 1- u 2 du = sin- 1 x- 3 2 + c, for x- 3 2 1 3. Z 4 x + 1 2 x 2 + 4 x + 10 dx Solution: This one requires a bit of thought. Note that the derivative of the denominator is 4 x + 4, which is almost the same as the numerator. Thus we add and subtract 3 from the numerator: Z 4 x + 1 2 x 2 + 4 x + 10 dx = Z 4 x + 4 2 x 2 + 4 x + 10 dx- Z 3 2 x 2 + 4 x + 10 dx For the first integral: let u = 2 x 2 + 4 x + 10 du = 4 x + 4 and we have Z 4 x + 4 2 x 2 + 4 x + 10 dx = Z du u = ln 2 x 2 + 4 x + 10 . For the second integral, complete the square in the denominator: 2 x 2 + 4 x + 10 = 2( x 2 + 2 x + 1) + 10- 2 = 2( x + 1) 2 + 8 Page 2 and so- Z 3 2 x 2 + 4 x + 10 dx =- 3 Z 1 2( x + 1) 2 + 8 dx =- 3 8 Z 1 ( x +1 2 ) 2 + 1 dx Now let u = x + 1 2 du = 1 2 dx and we find that- 3 8 Z 1 ( x +1 2 ) 2 + 1 dx =- 3 4 Z du u 2 + 1 =- 3 4 tan- 1 x + 1 2 . Putting it all together, Z 4 x + 1 2 x 2 + 4 x + 10 dx = ln 2 x 2 + 4 x + 10- 3 4 tan- 1 x + 1 2 + c. (This is an indefinite integral, so we must remember to include the constant of integration, c .) 4. Z ( x 2- 5) 2 dx Solution: Expand the integrand: Z ( x 2- 5) 2 dx = Z ( x 4- 10 x 2 + 25) dx = 1 5 x 5- 10 3 x 3 + 25 x + c. 5. Z 1 e x 1- e 2 x dx Page 3 Solution: Let u = e x du = e x dx Thus Z 1 e x 1- e 2 x dx = Z e 1 du 1- u 2 = sin- 1 ( e )- sin- 1 (1) = sin- 1 ( e )- 2 ....
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IntegrationReviewSolns - Integration Review MATH 213-C...

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