You can still use the first reverse chain rule ‘trick’ when the function is multiplied by a constant, k.
Transcript
Last time, we saw how to integrate fractions where the numerator is the derivative of the denominator.
For example, the integral of this fraction is… [5ex+2x5ex+x2dx]
The integral of this fraction is this [ln(5ex+x2)+c ]
Now, we’ve seen that this trick does not work for a fraction like this [5ex+25ex+x2dx, cont. conv.]…
As it only works when this expression is the derivative of this one…
However, we can extend this trick a little further than we have so far… [back to 5ex+2x5ex+x2dx]
We can also use it in cases where this expression is any multiple of the derivative [now to 3(5ex+2x)5ex+x2dx]
And that’s because it’s possible to take this coefficient [3]…
And write it here instead [35ex+2x5ex+x2dx]
Allowing us to evaluate this integral with the trick [above]… [click one time after animation finishes]
Evaluating this integral gives this, and the 3 stays where it is! [3ln 5ex+x2 +c]
And the derivative on top could also be multiplied by a non-integer…
Like here, for example… [653cosx +53sinx +5x dx]
Given that this [3cosx +5] is the derivative of this [3sinx +5x], evaluate this integral…
First, we can write this coefficient [65]…here [653cosx +53sinx +5x dx]
And second, we can evaluate this integral to get this [65ln 3sin x +5x +c]
So, the first integration trick also works when the top of a fraction is any multiple of the bottom’s derivative…
However, in each of those last two examples, the top of the fraction was conveniently factorised…
Show 3(5ex+2x)5ex+x2dx and65(3cosx +53sinx +5x dx
So that we had the derivative in brackets… [highlight in examples]
And then some coefficient outside the brackets [highlight]
And that made it easy to see that these expressions are multiples of the derivatives! [whole expressions on top]
But what if the tops of these fractions weren’t factorised?
Show 15ex+6x5ex+x2dx and185cos x +63sinx +5x dx
Well, it would then be quite a lot harder to tell that these expressions on top, are multiples of these expressions’ derivatives….
But, they still would be!
So, if we want to use the first integration trick to the best of its ability…
We need to be able to see fractions like these…which aren’t factorised…
And recognise that these expressions…[top]…
Are multiples of the derivatives of these expressions [bottom]…
So, we’ll look at how to do that….next…
But to sum up for now, the first integration trick also works when the top of a fraction…
Is a multiple of the bottom’s derivative…
For example, the integral of this general function [kf’xfxdx], is…
The integral of this general function is, first, the same as this [kf’xfxdx]
Meaning it’s then…this [kln fx +c]