5.5.1-eg-1 - Example Evaluate the indenite integral 6x 3...

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Example Evaluate the indefinite integral Z 6 x 3 5 x 4 + 12 dx. Solution: This problem is made difficult because of the square root of a function of x in the denominator. However, a simplification occurs when we make an appropriate substitution. Let u = u ( x ) denote the function of x under the square root, i.e., u = 5 x 4 + 12 , (1) and note that the differential du satisfies du = 20 x 3 dx. (2) The reason that u is a good substitution in this problem is that, after replacing the expression 5 x 4 + 12 in the integrand by u , the remaining x -dependent factor, x 3 dx , can be seen to be a constant multiple of du given above. That is, we can divide through by 20 in (2) to obtain x 3 dx = 1 20 du, (3) and obtain a simplification of the original integral occurs when the substitutions (1) and (3)
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