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MTH 102 MTH 102 Past Questions Objective Question

Question

Evaluate ∫ (e2x + e-2x)/(e2x - e-2x) dx
Options
A <i>(1/4)ln|e<sup>2x</sup> - e<sup>-2x</sup>| + C</i>
B <i>(1/2)ln|e<sup>2x</sup> + e<sup>-2x</sup>| + C</i>
C <i>(1/2)ln|e<sup>2x</sup> - e<sup>-2x</sup>| + C</i>
Correct Answer
D <i>(1/4)ln|e<sup>2x</sup> + e<sup>-2x</sup>| + C</i>
Correct Answer

Option C is the correct answer.

Detailed Explanation

This integral can be solved using u-substitution.Let u = e2x - e-2x.Then, calculate du/dx:du/dx = d/dx (e2x) - d/dx (e-2x)du/dx = 2e2x - (-2e-2x) = 2e2x + 2e-2x = 2(e2x + e-2x).So, du = 2(e2x + e-2x) dx, which means (e2x + e-2x) dx = (1/2) du.Substitute u and du into the integral:∫ (1/u) ∙ (1/2) du = (1/2) ∫ (1/u) du= (1/2) ln|u| + C.Substitute back u = e2x - e-2x:= (1/2) ln|e2x - e-2x| + C. This matches option C.

Hint

Look for a function in the denominator whose derivative is related to the numerator. This suggests u-substitution. Let u be the denominator.

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