sol16-3 - SOLUTIONS CHAPTER 16.3 MATH 132 - WI00 4. (3x2 +...

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MATH 132 - WI00 4. R (3 x 2 +14 x )( x 3 +7 x 2 +1) dx . Having two polinomials, take the one of bigger degree, in our case x 3 +7 x 2 +1, to be u and see if one can use it as the new variable: u = x 3 + 7 x 2 + 1 Find du by di±erentiating the formula in x : du = ( x 3 + 7 x 2 + 1) 0 dx = (3 x 2 + 7 * 2 x ) dx = (3 x 2 + 14 x ) dx Solve for dx : dx = du 3 x 2 + 14 x Try to use it in the integral: Z (3 x 2 + 14 x )( x 3 + 7 x 2 + 1) dx Z (3 x 2 + 14 x ) u du 3 x 2 + 14 x Z udu = u 2 2 + C = ( x 3 + 7 x 2 + 1) 2 2 + C 14. Z x p 1 + 2 x 2 dx Again, two polinomials (one is x and the other one is 1+2 x 2 ), so choose the one of higher degree to be u u = 1 + 2 x 2 . Let’s see if it fits: du = (1 + 2 x 2 ) 0 dx = (0 + 2 * 2 x ) dx = 4 xdx Solve for dx : dx = du 4 x Z x p 1 + 2 x 2 dx = Z x u du 4 x Z u 1 4 du = 1 4 Z u 1 2 du 1 4 u 3 2 3 2 + C = 1 4 (1 + 2 x 2 ) 3 2 3 2 + C 26. Z
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sol16-3 - SOLUTIONS CHAPTER 16.3 MATH 132 - WI00 4. (3x2 +...

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