Evaluate the integrals in Exercises 9–28. It may be necessary ...

Question

Evaluate the integrals in Exercises 9–28. It may be necessary to use a substitution first.

∫ [1 / √(e^s + 1)] ds

Accepted Answer

Hello there. Today we are going to solve the following practice problem together. So first off, let us read the problem and highlight all the key pieces of information that we need to use in order to solve this problem. Evaluate the integral, the integral of 1 divided by the square root of e to the power of x + 4 dx. OK. So it appears for this particular problem, we're asked to simply evaluate the integral that is provided to us. So now that we know that we're ultimately trying to evaluate this specific integral, our first step that we need to take is we need to know, based on what we can see in our integral, we can recall and use the substitution method. So let us say that you, for this particular prong, will be equal to the square root of e to the power of X + 4. Fantastic. And we need to note that this is the most often the direct path for integrals, or the most direct solving method path for integrals that involve a root of an exponential, specifically. So now, we need to solve for DX and in order to find DX we need to square both sides, we'll find that U 2 is equal toe to the power of X + 4. And now we need to isolate and solve for E of X, so E of X is equal to U 2 minus 4. And now we need to differentiate, so E to the power of XDX is equal to 2 UDU. And now we need to replace e to the power of x, so we need to write parentheses of u to the 2 minus 4 dx is equal to 2UDU, which will mean that dx is equal to 2U divided by u to the 2 minus 4DU. So now we need to substitute our values or specifically these values back into our integral, so we could write that I where I represents our integral, so I equals. The integral of 1 divided by u multiplied by 2 multiplied by u divided by u to the power 2 minus 4DU, and we need to note that the u terms will cancel out, so you can go ahead and write that I will be equal to the integral of 2 divided by u to the 2 minus 4DU. Fantastic. So now, our second step now is we need to recall and use partial fraction decomposition. So we need to note for this particular problem, the integrand is 2 divided by u to the 2 minus 4. And we need to note that this can be factored as 2 divided by parentheses of U minus 2 in parentheses, multiplied by parentheses of U + 2. Fantastic. And we can split this into partial fractions. So, we can say that 2 divided by parentheses U minus 2 parentheses multiplied by parentheses of U + 2 will be equal to a divided by U minus 2, plus B, which is gonna be divided by U plus 2. So now, our next step, and let's make a note off to the side, we need to multiply both sides of our expression by the denominator, which is 2 is equal to. So when we multiply by the denominator, we will be able to write that 2 will be equal to. A multiplied by parentheses of U + 2, so U + 2 in parentheses, plus B multiplied by parentheses of U minus 2. Fantastic. In keeping that expression in mind, we can set u equal to 2, which will mean that 2 is equal to 4 multiplied by A, and when we rearrange this expression to isolate and solve for A, we will find that A is simply equal to 1/2. And when we set U is equal to 2, we'll be able to write that 2 is equal to -4 multiplied by B, and when we rearrange and isolate and solve for B, using a little bit of algebra, we will find that B is equal to 1/2. Which will mean our integral will become i is equal to the integral of parentheses of 1/2 divided by u minus 2. -1/2 divided by you plus 2. DU. Fantastic. So moving right along here, so that will mean that we can go ahead and move on to step number 3. And step 3 is we need to begin our integration, so we need to integrate the terms. So we could state that I is equal to 1/2 multiplied by the natural log of the absolute value of u minus 2, and don't forget your absolute value signs minus 1/2 multiplied by the natural log of the absolute value of u plus 2. Plus c or c denotes some constant variable, so we now at this stage need to recall and use the logarithic property that the natural log of a minus the natural log of b is equal to the natural log of a divided by b. So using that property, we can go ahead and write that i is equal to 1/2 multiplied by the natural log of the absolute value of u minus 2 divided by U + 2. In absolute value signs plus C. Fantastic. So our fourth and final step is we need to use back substitution. So we need to replace U, and we need to note that U is equal to the square root of E to the power of X + 4. So that means we could write that i is equal to 1/2 multiplied by the natural log of the absolute value of the square root of e to the power of x + 4. And this will be -2. All divided by the square root of e to the power of x + 4 + 2. Plus c and once again c denotes some constant variable and that's it we've officially solved for this problem hooray, we did it. So going back to the top to look at the prom itself we could safely say once again without a doubt. That our final answer for the evaluated integral is as follows 1/2 multiplied by the natural log of the absolute value of the square root of e to the power of x + 4 minus 2 divided by the square root of e to the power of x + 4 + 2 + c. And that's it, that is our final answer. Thank you so much for watching, hopefully that helped, and I can't wait to see you in the next video. Bye.

Evaluate the integrals in Exercises 9–28. It may be necessary to use a substitution first.∫ [1 / √(e^s + 1)] ds

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Evaluate the integrals in Exercises 9–28. It may be necessary to use a substitution first.
∫ [1 / √(e^s + 1)] ds