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Verify by differentiation that the formula is correct.$\displaystyle \int \frac{1}{x^2 \sqrt{1 + x^2}} \,dx = - \frac{\sqrt{1 + x^2}}{x} + C$

$\frac{1}{x^{2} \sqrt{1+x^{2}}}$

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Kayleah T.

Harvey Mudd College

Samuel H.

University of Nottingham

Michael J.

Idaho State University

Boston College

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Video Transcript

All right. What they're asking us to do is take the derivative of the answer. That negative square root of one plus X squared all over X plus C. And double check that at equal. So, what we have going on is the quotient rule. So the derivative of the top is actually the chain rule inside of that. Which I imagine if you're at this point then you should be good. Um One plus X squared is now to the negative one half. Power times the derivative of the inside shit times you leave the bottom alone minus the derivative of the bottom, which is just one. You leave the tap alone, which I'll leave it as that. Uhh make it plus square root of one plus X squared. All over the denominator squared in the derivative of a constant zero. Now, at this point, we might have some magic happened to cancel things out The 1/2 and the two will cancel. Um So what I'm looking at is the negative of let's see this kind of gross as one over bloody X squared over square root of one plus X squared. This is still plus the square root of one plus X squared. All over X squared. Now I can get the same denominator and then numerator and sounds weird by just multiplying by the square root of one plus X squared. Um And the reason why you would do that is now this and this can cancel, you'd have negative X squared with a later positive X squared. So in the top you would be left with one and these two pieces we could combine together and stick into the denominator. And let me put it this way that dividing by X squared is the same thing as multiplying the top one over X squared. Uh And I believe that's what we started with. So our whole goal was to get to this piece. So I mean circle it in green and that's what they wanted you to do to take the derivative and get it equal to the other side. So we did it.

Topics

Integrals

Integration

Kayleah T.

Harvey Mudd College

Samuel H.

University of Nottingham

Michael J.

Idaho State University

Boston College

Lectures

Join Bootcamp