evaluate-the-integrals-by-making-appropriate-u-substitutions-and-applying-the-formulas-reviewed-i-55

Question

Evaluate the integrals by making appropriate $u$ -substitutions and applying the formulas reviewed in this section.
$$
\int \frac{e^{x}}{\sqrt{1-e^{2 x}}} d x
$$

Julian Gerber · Accepted Answer

So here we have the integral of eat of the X over the square root of one minus eat of the two X. So to solve this, we&#x27;re gonna use the use substitution where U equals e to the X, which means that d you is just either the axe DX. It means we can instead of the numerator e to the x dx, we&#x27;d just sub student d&#x27;you over the square root of one minus. You squared. Now we&#x27;re gonna use a trigger substitution where instead of you we&#x27;re gonna set you equal to sign of Z. Let&#x27;s say which means that, um d u equals, um, the co sign of Z d Z And the reason we&#x27;re doing this is ultimately we&#x27;re going to use the identity one minus sine squared. X equals co sine squared X. If we get if we can use that identity, things become very easy. So instead of d u in the numerator we have co sign Z DZ over square root of one minus sine squared Z one minus sine squared Z We&#x27;ve already established its co sine squared of Z. Now we have co signs e TZ over the square root of co sine squared, which is just co signs E Now we see these cancel out. So all we&#x27;re left with is the integral of DZ, which we know from the internal table is Jay Z plus the constant of integration. Now, um, we have to change this, z, um, back into you and then ultimately back into in terms of X. So we have you equal sine z, which means that Z equals inverse sign of you so this can become in verse sign of you plus c you we already know is e to the X. So all this becomes is inverse sign of e to the X plus the constant of integration.

Geena Pullo · Answer

Hi. So this question we&#x27;re going to be solving for the indefinite integral using the method of you substitution. So the first thing I&#x27;m going to do is I&#x27;m just going to rewrite our equation as sex. Sorry, Um, as e v to the act over thesis with root of one squared minus E to the X squared DX. This might seem a little odd, but it&#x27;ll make more sense is we continue. And from here, we need to do a U substitution. So we&#x27;re going to set Are you equal to e to the X? And then we&#x27;re gonna take the derivative of both sides. So we&#x27;re gonna have being you is equal to be to the x d x. And now from here, what we&#x27;re going to do is we&#x27;re gonna plug in for our each acts Are you were going to plug in our D you in for r e to the x dx. So we&#x27;re just gonna be great. And it&#x27;s going to now be R D you over the square root of one squared minus, you squared. And so from here, we can actually solve our, uh, integral using this equation. So we have the equation that suits that the integral of D you over the square root of a squared minus you squared is equal to the ark sign of you over a plus, e. So we&#x27;re now going to just plug in for our value. So we&#x27;re gonna have our one. It&#x27;s equal to our A And are you in this case, is, of course was equal to you. So we&#x27;re going to have the and a girl. I&#x27;m not the interest that really that And we&#x27;re gonna have the are something of you over one plus see. And from here, all we have to do is plug in for our view and our u t too to the X, right? Yep. And so our answer is going to be these arc sine of e to the X plus. See, that is the final answer.

Geena Pullo · Answer

So in this question, we&#x27;re going to be solving the indefinite integral. Using the method of the substitution, we have the integral of E to the square root of X minus one over the square root of X minus one. So first thing we&#x27;re going to do is we&#x27;re going to set. Are you equal to the square root of X minus one? And I we&#x27;re gonna take the derivative of both sides. And so the derivative of the square root of X minus one. What we&#x27;re going to do for that, just as a reminder is, we&#x27;re going to take both. We&#x27;re going to the powerful and the chain room. So this is going to be equal to X minus one to the one half, and from here, we&#x27;re going tohave one half times X minus one. Uh, yes. Times X minus one to the one half minus one. And this is going to be times our internal derivative of just one. So we&#x27;re now going tohave one half times X minus one to the negative one half, which is just going to be equal to 1/2 times the square root of X minus one. De looks I&#x27;ll put a d x here. Just a reminder. And so from here, what we&#x27;re going to do is we&#x27;re going to replace R square root of the X minus one with argue and r d x times X minus one with r D r two d u Because we&#x27;re gonna bring are gonna multiply our two times are D&#x27;You um, and times both sides really to get it out of our denominator here and now we can be right our equation. So we&#x27;re just going to have the integral of e to the square root. I have, uh, not square root. Sorry. Each of the U I am two d u times to do you. I am so from here we can take out Are you? I was just going to have the integral of e to the U D you and the integral of each of the u. D. You is just equal to each of the you, so we&#x27;re going to have to times e to the U plus r c on from here. We have to just fill in for our you and are you is equal to the square root of X minus one right And so we&#x27;re going to just have to times e to the square root of X minus one. Let&#x27;s see, and that is our final answer.

Julian Gerber · Answer

So here we have the integral of e to the square root of X plus one divided by the square root of X plus one. Now Teoh, solve this. We&#x27;re gonna use the U substitution. U equals square root X plus one which, remember that&#x27;s the same thing is rating X plus one to the one half. That means D you using the power rule and the chain rule is one half X plus one to the negative one half times the derivative of X plus one, which is just one so we can leave that out now. This is the same as writing, um, one over two or one half times square root X plus mourn d X. Now we have a square root X plus one in the denominator. Um, so all we have to do is include a one half there. If we&#x27;re going Teoh, multiply the inside of the integral by one half. We have to multiply the outside by two to keep things balanced. So now we have. If we substitute, we have to times the integral of e to the U. D. You now the integral of eat of the U is just eat of the years we have eat two times e to the U plus the constant of integration. Now all we have to dio is substitute in square root X plus one back in for you. So we have two times e to the square root of X plus one, plus the constant of integration.

Geena Pullo · Answer

I was solving for the indefinite Integral Using you substitution. The first thing we&#x27;re going to be doing is we&#x27;re gonna just change the way this integral looks a little bit. We&#x27;re gonna manipulate it. So we&#x27;re gonna have the integral of E to the X over the square root of to squared minus E to the X squared. Because either the X squared is equal to e to the two acts Ah d x. And from here, we&#x27;re going to do a u substitution. So we&#x27;re gonna have you is equal to e to the X. And now we&#x27;re gonna take the derivative of both sides when a d. U is equal to e to the x t x Andi Um, what we&#x27;re going to do is we&#x27;re just going to replace R E to the X DX with our Dumont replace R E to the acts with Are you so now we&#x27;re going tohave integral of D u over the square root of two squared minus you squared. And from here, we&#x27;re going to do is we&#x27;re gonna be using on equation to solve this. So there&#x27;s an equation that states that the integral of d U over the square root of a squared minus you squared is equal to the ark Sign of you over a plus C And so we&#x27;re going to have art A be equal to two And are you equal to you in this case? So first thing we&#x27;re gonna do is we&#x27;re going to rewrite our equation. Having solved are integral. So we have the sign, the arc sine of you over to plus C. And now from here, all that is left to do is to, uh, re put in r E to the X for Are you so we&#x27;re going tohave be sign the arc sine of e to the X over two plus C and that is our final answer.

Julian Gerber · Answer

So here we have the integral of eat of the X over the square root of four minus eat of two x dx. Now, to solve this, we&#x27;re gonna use the use substitution u equals E to the X and the derivative that is just again e to the X. So if we substitute, we can substitute the numerator for D U over square root four minus you squared. Now, luckily for us in the integral table, there is a formula for integral of the form D you over square root a square to minus you squared And that is just in verse sign of you over a plus the constant of integration. So in this case, a squared is four So a is to because we only look at the positive square room We need a to be positive. So using this formula, all we get is the inverse sign of you over two plus c now, substituting either the ex back in for you. We get in verse sign of e t X over two plus the constant of integration

Problem

Evaluate the integrals by making appropriate $u$ …

Problem 25 Easy Difficulty

Evaluate the integrals by making appropriate $u$ -substitutions and applying the formulas reviewed in this section.
$$
\int \frac{e^{x}}{\sqrt{1-e^{2 x}}} d x
$$

Answer

$\int \frac{e^{x}}{\sqrt{1-e^{2 x}}} d x=\arcsin \left(e^{x}\right)+C$

Related Courses

Calculus Early Transcendentals

Related Topics

Discussion

Top Calculus 2 / BC Educators

Grace H.

Anna Marie V.

Caleb E.

Kristen K.

Calculus 2 / BC Courses

Integration Review - Intro

Definite vs Indefinite

Recommended Videos

Watch More Solved Questions in Chapter 7

Video Transcript

Howard Anton, Irl Bivens, Stephen Davis

Calculus Early Transcendentals

Grace H.

Anna Marie V.

Caleb E.

Kristen K.

Integration Review - Intro

Definite vs Indefinite

What do you need help with?

Textbooks

Ask a question

Courses

AI Tutor

$\int \frac{e^{x}}{\sqrt{1-e^{2 x}}} d x=\arcsin \left(e^{x}\right)+C$

Calculus Early Transcendentals

Grace H.

Anna Marie V.

Caleb E.

Kristen K.

Integration Review - Intro

Definite vs Indefinite

Howard Anton, Irl Bivens, Stephen DavisCalculus Early Transcendentals

Grace H.

Anna Marie V.

Caleb E.

Kristen K.

Howard Anton, Irl Bivens, Stephen Davis

Calculus Early Transcendentals