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Numerade Educator



Problem 40 Hard Difficulty

Make a substitution to express the integrand as a rational function and then evaluate the integral.

$ \displaystyle \int \frac{dx}{2 \sqrt{x + 3} + x} $


$\int_{\text { the solution. }} \frac{d x}{2 \sqrt{x+3}+x}=\frac{1}{2} \ln |\sqrt{x+3}-1|+\frac{3}{2} \ln |\sqrt{x+3}+3|+C .$ Click to see


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Video Transcript

let's star this problem by using a substitution. And then after the substitution, we should be able to write this fraction as a rational function. Let's take you to be that radical and the denominator. That's radical. Excl Astri. Then we have to you one over two with that radical expose three. And here we congrats. This as one over to you, d X because U equals a radical. And then I could multiply both sides by two. You and then I have this equation here. This tells me that I could come over here to the general and replaced the ex with to deal in the denominator. This is also a to you and I don't want to write X here because I have a new variable You. So I have to come to this substitution in Software X and there it is. So, exes, you square minus three. No. So, also, for this inner growth would like we can rewrite this, Maybe pull up that too, whether you up top you square plus two U minus three. So we do have a rational function, and then we can let's go over the side and set up the partial fraction the composition. You're up top. You plus three. You minus one in the bottom. This is what the book calls Case one. We have two distinct living. Your factors in the denominator sort of a over. You plus three be over. You minus one. Just go ahead and multiply both sides of this equation. By this term, on the bottom left, we get you equals air times you put minus one plus me, you clustering. And then here we could fact they're out. Ah, you and that are constant. Term is three B minus a. All right, so now let's look on the left and farming to you. We see that there's no one there. That means a plus B is one. We don't see a constant storm on the left. That means it must be zero. So we have three a minus B equals zero. Excuse me. That should have been a three B minus a wrong letters there. So we have a equals three B plug that into the first equation. You get four b equals one. So be is one fourth, and then we have a equals three fourths. So let's just go ahead and plug in these and be values into our partial fraction the composition. And then well, you never know. So I'm running out of room here and let me go to the next stage. So don't forget the two that we originally had outside the integral. And then we have three over fourths. That was our a overyou clustering. And then we had one of her fourth. That was R B overyou minus one, and we're ready to integrate. If this plus three is bothering you, feel free to do it. Use up. Or in this case, let me different letter. Maybe a w substitution over here if that minus one is bothering you. Similar idea. Doing these substitution Sze we should get We have two times threw you over for that's three over to natural log you plus three in the absolute value. Plus two times a fourth. That's a half natural log. Absolute value, Hugh. Minus one Plus he. And here the last thing to do would be to go ahead and right this back in terms of the variable X. So this is all equal to three half natural log and then you was the radical X plus three It's not necessary to use absolute value here because a radical is bigger than your equals. Zero There's a three. So this thing is always positive, so you could lose the absolute value if you want. Then we have not one half natural odd absolute value Ex close three minus one plus our constancy and that's our into.