


IntegrationIntegration by Substitution 2  Harder Algebraic SubstitutionWe assume that you are familiar with the material in integration by substitution 1. Recall the Substitution Rule To integrate if we replaceby andby. This converts the original integral into a simpler one. We extend this idea to some harder integrals. Sometimes an integrand may need a bit of algebraic manipulation to make it integrable. Make sure that the integral with respect to the variable u does not contain any x. Consider some special cases: f(ax^{2} + b)x^{m} dxf(ax^{2} + b)x^{m} dx where m is odd can be rewriiten as f(ax^{2} + b)x^{m − 1} x dx and the substitution u = ax^{2} + b used. ExampleLet us consider x^{3}(x^{2} + 1)^{99}dx. (x^{2} + 1)^{99} is a composite function. One would consider replacing the inner function (x^{2} + 1) by u. Thus du = 2x dx, which can be obtained from x^{3} since it equals x^{2} times x. We rewrite the integral as ½x^{2}(x^{2} + 1) ^{99} 2x dx. Since u = x^{2} + 1, x^{2} = u − 1 Hence the integral becomes ½(u − 1)u^{99} du = ½(u^{100} − u^{99}) du This is integrable. ½(u^{100} − u^{99}) du = ½((1/101)u^{101} −(1/100)u^{100}) + c = (1/202)(x^{2} + 1)^{101} − (1/200)(x^{2} + 1)^{100} + c ExerciseIntegrand contains (mx + b)^{(m/n)}When the integrand contains an expression of the form (mx + b)^{(m/n)} then the substitution u = (mx + b)^{(1/n)} is often suitable. Thus u^{n} = mx + b and nu^{n − 1} du = m dx. ExampleEvaluate . The square root is a composite function with inner function 2x + 1. Let u = which is equivalent to u^{2} = 2x + 1 Then 2u du = 2 dx or u du = dx. We need to express the factor x in terms of u. u^{2} = 2x + 1 so x = ½(u^{2} − 1) We can write the integral as Hence the integral becomes Integrating and substituting for u:
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