5.5.1-eg-1 - Example Evaluate the indefinite integral √...

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Unformatted text preview: Example Evaluate the indefinite integral √ 6x 3 5x4 + 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 = 5x4 + 12, (1) and note that the differential du satisfies du = 20x3 dx. (2) The √ reason that u is a good substitution in this problem is that, after replacing the expression √ 5x4 + 12 in the integrand by u, the remaining x-dependent factor, x3 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 1 (3) x3 dx = du, 20 and obtain a simplification of the original integral occurs when the substitutions (1) and (3) are made; i.e., 6x3 dx = 6 3x4 + 12 1 x3 dx + 12 1 1 =6 √· du u 20 3 −1/2 u du, = 10 an indefinite integral in the new variable of integration u, and one for which the solution is easily found. Thus √ √ √ 3x4 6x3 3 u1/2 dx = + C. 10 1/2 3x4 + 12 Now replacing u by 5x4 + 12 in the solution given above, it follows that 6x3 3 (5x4 + 12)1/2 dx = +C 10 1/2 3x4 + 12 3√ 4 = 5x + 12 + C, 5 so that the final result is entirely in terms of x, the original variable of integration. √ ...
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