7–84. Evaluate the following integrals.
33. ∫ [eˣ / (a² ...

Question

7–84. Evaluate the following integrals.

33. ∫ [eˣ / (a² + e²ˣ)] dx, where a ≠ 0

Accepted Answer

Welcome back, everyone. Evaluate the integral. Integral of 2 E's the power of X divided by 9 + 4 eats the power of 2 XD X. So let's begin solving this problem by performing factorization. First of all, we can factor out 2 from the numerator. Within the integral we're going to have integral of e to the power of x divided by 9 + 4 e to the power of 2 X followed by D X. And also let's factor out 9 in the denominators. So we get 2 integral eats the power of X D X divided by 9 in 1 + 4 divided by 9 eats the power of 2 X. We can also take out 9, so that it becomes a constant in the denominator. Meaning we get 2 divided by 9 integral, eats the power of XD X. Divided by one plus. 4 divided by 9 e to the power of 2 X. Now, notice that 4 divided by 9 can be written as 2/3 squared. So we can also use that idea and rewrite our integral as 2 divided by 9 integral eats the power of XDX. Divided by 1 plus. 2/3 of the power of X all squared, right? Because 2/3 squared gives us 4 divided by 9, and eats the power of X squared this eats the power of 2 X. And then we can clearly see that one of the substitution methods that we can use to begin with is going to be U equals e to the power of X. So the U divided by D X is the derivative of eats the power of X, which is eats the power of X, and we clearly see that E to the power of X D X is equal to DU, right? If we multiply both sides by D X. So our integral becomes 2 divided by 9 integral. Of the you divided by. One plus. 2/3U squared. And from here we can see that this form. It's very similar to the integral that gives us inverse of tangent. So, what we're going to do is simply introduce another substitution. Let's suppose that V equals 2/3 U. Meaning DV. Divided by the EU is equal to 2/3. And therefore, we can show that the U is equal to DV multiplied by 3 divided by 2, so 3 halves DV. So, in terms of V, our integral becomes 2 divided by 9, integral of DU, which is 3 halves DV, we can just write DV and we can factor out three halves, right? So we get 2 divided by 9, multiplied by 3 halves, integral DV divided by 1 plus V squad. And that we know that this integral is equal to inverse of tangent. So, what we get is, first of all, our constant 2 multiplied by 3 is 6, and 6 is divided by the product between 9 and 2, right? So, 6 divided by 18 is 13. Followed by inverse of tangents. Off V plus a constant of integration C. And now we are almost done. We just want to substitute for V, that'd be 1/3 inverse of tangents. Of 2/3 u plus C. And finally, we want to get our expression in terms of X. So that'd be 1/3 inverse of tangent. Of 2/3, and now U was E to the power of X, so we get 2/3 E to the power of X, and finally plus C. And that'll be our final answer for this problem. Thank you for watching.

7–84. Evaluate the following integrals.
33. ∫ [eˣ / (a² + e²ˣ)] dx, where a ≠ 0

Key Concepts

Integration Techniques

Exponential Functions

Improper Integrals

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