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# Make a substitution to express the integrand as a rational function and then evaluate the integral.$\displaystyle \int \frac{e^x}{(e^x - 2)(e^{2x} + 1)}\ dx$

## $\frac{1}{5} \ln \left|e^{x}-2\right|-\frac{1}{10} \ln \left(e^{2 x}+1\right)-\frac{2}{5} \arctan e^{x}+C$

#### Topics

Integration Techniques

### Discussion

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##### Kristen K.

University of Michigan - Ann Arbor

##### Samuel H.

University of Nottingham

##### Michael J.

Idaho State University

Lectures

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### Video Transcript

let's make a substitution to rewrite this in a grand as a rational function. Looking at the numerator, we should just take you to be either of X, sir. Then do you is just either the ex dx so we can rewrite this integral as well. The numerator This is to you now in the denominator that becomes you minus two and then we have eat of the two X there. So either the two X it's just eat of the X squared, which is use where? So we have you squared plus one. Then we have to check whether or not this quadratic faster will factor into two in years. So to do this, you look at the discriminating B squared minus four A. C and our problem be a zero because there's no u turn here. So zero squared minus four A is one, and then C is one. So we had a negative quantity. That means that this quadratic does not factor. So, using case one in case three in your textbook. Oh shit. And then for the second fraction, bu plus he and then we should go ahead in the side and actually find a B and C. So, really, what we should be having here is let's ignore the integral For a moment we have won over you might ooh, you squared plus one by a previous reasoning. Multiply both sides of this by the denominator on the left, and that's what we get on the right. And then we just go ahead and simplifying that ready inside a squared plus a B square. See you minus two BU minus two C and then we can go ahead and combined like terms. So let's pull a you square, we get a plus B, Let's pull out of you. And when we pull, are you We have C minus two B and then a minus to see. Okay, so now we matched the coefficients on the left in the rake on the right. Actually, let me start on the left. We see that the constant term is the one that means a minus two C is one. And since there's no you were you squared on the west, that must mean that a plus B and then C minus to me or both, zeroes less right and solve the system On the next page, a minus two c equals one. So we also had a plus B equals zero and then C minus two equals zero. So, for example, here we have it was negative. Here we have c equals to be which we can run his negative to me from this equation here. And so coming back to this equation of here we have a minus to see. But then, see is just negative to a So this becomes a plus two times two equals one. So we get a equals one fifth, then from this equation here he is negative. And from this equation here she is negative too. Plugging these back into the integral in the first page after plugging for a B and C that's you, minus two down there. But then plus and then minus one over five and then times you minus two over five. That was our being sin terms use were plus one, Do you? Now we can go out and split this into three intervals. So it's right that next actually, let me go to the next page here I'Ll be more room. We pull out the confidence and split up along these inner girls as much like you. That's the first one. Then I pull out a negative one fifth from the second one. You over You squared plus one and then pull up the minus two over five. From the last fraction one of you squared plus one mom for the first rule. If this minus to his mother in here your u substitution, You'LL get one over five Natural. Are you minus two for the first internal? Then for the second, you can do it yourself here as well. So what? B squared plus one dw over too. It's just you. So we have one over w natural on w and then back substitute. And if you want, you could drop the absolute value here because you swear plus one is positive. So then also multiplied by this negative one of the five. So we get negative one over ten natural law you squared plus one And then for this loss interval we just memorize this as are you Now If you have forgotten that fact, that's perfectly fine in the way that you would evaluate. This is we would have to go to the side and the truth substitution, you know, and must you simplify you'LL end up with tannin burst data. We have your constancy. The last step here is to replace you with X So what, she's going to do that? No one that's either the ex. That was how we define you. And then here I should keep my absolute values there minus one over. Oh, that and then need to the two X plus one and then minus two over five CNN verse, you know, the ex plus RC, and that's our entry.

JH

#### Topics

Integration Techniques

##### Kristen K.

University of Michigan - Ann Arbor

##### Samuel H.

University of Nottingham

##### Michael J.

Idaho State University

Lectures

Join Bootcamp