Original subtitles

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Today we're going to be talking about how to use substitution to evaluate an indefinite integral.

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And in this particular problem we've been given the integral of 4 X divided by the square root of x

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squared plus 1.

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Now we've been told already that we need to use use substitution to evaluate this integral.

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But even if we hadn't been told that specifically that would be the first thing that we want to try

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to use to evaluate this integral and the reason is because we have kind of an obvious giveaway for you

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substitution or at least for trying it which is we have two values of x and our integral and they differ

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by one degree and what I mean by that is we have this x value in the numerator and this is essentially

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X to the first power right because it's just this a single value of x Here x x to the first power.

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We also have in our denominator here X squared or X to the second power and therefore in our numerator

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we have a first degree x value in our denominator we have a second degree x value and so we have two

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different x values and they differ by 1 degree.

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When that's the case it's often very convenient to set you equal to the higher degree x value or the

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higher degree x value plus whatever goes with it.

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So in this particular case because we have x squared plus one under the square root sign it's probably

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going to be helpful for us to set you equal to x squared plus 1 because then we'll be able to put you

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underneath the square root sign.

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We'll just have the square root of you as opposed to the square root of you plus 1 which would be a

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little bit messier and harder to deal with.

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So that's one way that we can determine what we're going to set you equal to.

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Now remember that a huge substitution problem basically all we're doing here is using a substitution

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to simplify our integral so that it's easier to evaluate because for X divided by the square of X plus

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one is a little bit of a complicated integral it's not just some simple polynomial that we use.

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Power rule with to evaluate the integral.

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It's a little bit more complicated and we can't see very clearly how we're going to evaluate this integral

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as it is in its original state.

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So we want to try you substitution first because that's our quickest method to evaluate the integral

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if it works.

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If you substitution doesn't where we might want to try integration by parts and if that doesn't work

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we might want to consider partial fractions and then maybe some other methods after that.

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So we've identified for you this x squared plus one value.

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Once we identify this value for you which is the value we're going to substitute we need to take the

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derivative of you and the derivative of you.

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We're going to denote as d you and all we're going to do is take the derivative of the right hand side.

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So the derivative of x squared is just 2 x the derivative of 1 is zero.

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So we don't need to include that in our derivative here.

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We just have d equals to X and once we're done taking the derivative here we just multiply this right

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hand side here by d X. That's really important because it allows us to make a proper substitution.

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Later on in this integral.

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So once we have you NDU then we can start substituting back into our original integral we'll go head

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here and rewrite our integral.

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Keep for x in the numerator because we haven't identified a substitution value at all for 4 x so that's

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going to stay.

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But in our denominator we said that was equal to x squared plus 1.

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So we're going to take the square root of you instead of x squared plus 1 because we set x squared plus

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1 equal to you.

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So that's our denominator then for d x what we want to do is solve this equation we wrote down for you.

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We want to solve it for Dayaks So we're going to come back here like this and we're going to divide

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both sides by 2 X which will solve the right hand side for Dayaks and we're going to get you over to

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x is equal to x.

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Now we have a value for X that we can plug into our original integral So instead of putting d x in here

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to our substitution integral we're going to put in d u over to x.

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Now once we've gotten to this point we want to simplify the integral as much as we can.

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So as you can see we've got x to the first power and x to the first power in our numerator and denominator

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here.

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Those are going to cancel with one another.

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We also have four and two in our numerator and denominator respectively.

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That too is going to cancel.

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And this four here is going to become just 2 for over two is two we can pull that two out in front of

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our integral because it's a constant coefficient on this one over the square of you term here.

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And so what will be left with is two times the integral of one over the square root of you do you.

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Now from here the simplest way to evaluate the integral is to transform this fraction that we have right

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here 1 divided by the square of you into just a polynomial a power term.

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What we're going to do is we're going to recognize that this of you is equal to you to the one half

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power of the same thing as a squirt of you.

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So essentially we have one divided by you to the one half power.

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What we're going to do is we're going to move that you to the one half power to the numerator we're

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going to take it out of the denominator and move it to the numerator.

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When we do that the sign on the exponent is going to flip.

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And so what we're going to be left with is two times the integral of you to the negative one half power.

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Do you like this.

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Now it's really easy for us to take the integral because we can just use power rule like we have before

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with simple indefinite integrals.

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So I'll be left with here is two and then we're going to multiply by whatever we get when we evaluate

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the integral here.

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Remember that the first thing we'll do is add one to the exponent so we're going to get you to the negative

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1 1/2 plus 1 we're going to add 1 to the exponent.

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Then we're going to divide our coefficient by our new exponent so our coefficient currently is just

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1.

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We're going to divide by our new exponent which is negative 1 1/2 plus 1 like this.

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And then we're going to add to this our cost of integration plus C.

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And the reason that we don't have to add the C inside of the parentheses and therefore multiply it by

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the two is because we just need C to act as a placeholder constant for whatever constant may have been

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there before we took the derivative of our original function and got this for X divided by the square

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of X plus 1.

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So C can say outside here by itself.

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All we need to do now is simplify what we've got in the parentheses.

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So we're going to have two times one over negative one half plus one is positive one half sort and get

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one over positive.

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One half you the positive one half like this plus.

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See now you divided by one half is the same thing as 1 times 2 over 1.

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We just take the reciprocal of this fraction in the denominator and we multiply by the reciprocal.

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Instead of dividing by the original fraction so we get 1 times 2 over 1 you to the 1 1/2 plus C and

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now you can see we're going to get 2 times 2 you to the 1 1/2 plus C.

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And at this point we can multiply the twos together what we'll get is four and we'd have four you to

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the one half but remember that you to the 1 1/2 is equal to the square root of you it's the same thing.

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So we're going to get 4 times the square root of you plus see at this point we'd be done except that

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we want our answer back in terms of X since we started our original problem in terms of x.

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You always want to finish with the original variable that you started with.

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So we're going to back substitute for you.

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And we're going to plug in x squared plus 1 again.

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We're going to bring that back and we're going to say that our final answer is four times the square

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root of x squared plus 1 plus C to account for that constant of integration.

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And that's our final answer.

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That's how you use your substitution to evaluate an indefinite integral.