Up Learn – A Level MATHs (edexcel) – Calculus III
Integrating kf'(x) Divided by f(x): Part 1
You can still use the first reverse chain rule ‘trick’ when the function is multiplied by a constant, k.
More videos on Calculus III:
The Reverse Chain Rule: Summary
Identifying f'(x) Divided by f(x)
Integrating f'(x) Divided by f(x)
Integrating kf'(x) Divided by f(x): Part 1
Integrating Parametric Equations: Summary
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Calculus III
2. The One Algebraic Function that Got Away
3. The Purpose of the Modulus
4. What Went Wrong?
5. What We Know So Far
6. Integrating Exponential Functions
7. Integrating Trig Functions
8. Integrating sin(x)
9. Integrating cos(x)
10. Integrating tan(x)
11. Integrating cot(x)
12. Integrating cosec(x)
13. Integrating sec(x)
14. Integrating Four More Trig Functions
15. Integrating sec^2(x)
16. Integrating cosec^2(x)
17. Integrating cosec(x)cot(x)
18. Integrating sec(x)tan(x)
19. Revisiting Definite Integrals
20. The Area Under a Sine Curve
2. What We Can Differentiate
3. Liouville’s Vision
4. The Fate of Integration
5. Identifying f'(x) Divided by f(x)
6. Integrating f'(x) Divided by f(x)
7. Integrating kf'(x) Divided by f(x) Part 1
8. Identifying kf’x Divided by f(x)
9. Integrating kf'(x) Divided by f(x) Part 2
10. Integration and Partial Fractions
11. Identifying f'(x) Multiplied by (f(x))^n
12. Identifying kf'(x) multiplied by (f(x))^n
13. Integrating kf'(x) Multiplied by (f(x))^n
14. The Reverse Chain Rule Part 1
15. The Reverse Chain Rule Part 2
16. Difficulty with the Reverse Chain Rule
17. Making a Substitution
18. Converting the Infinitesimal
19. Integrating with Respect to u
20. Integration by Substitution
21. Non-Linear Substitutions
22. Speeding Up the Process
23. Substitutions Where x is the Subject
24. Implicitly Defined Substitutions
25. Finding Your Own Substitution
26. Pick the Expression That’s Been Raised to a Power
27. Why Did We Learn the First Two Tricks?
28. Converting the Boundaries of a Definite Integral
2. The One Up, One Down Game
3. Mastering the One Up, One Down Game
4. Changing the Rules of the Game
5. When Both Expressions are Algebraic
6. When One Expression is Logarithmic
7. Integration by Parts 1
8. Integration by Parts 2
9. Completing Integration by Parts
10. Integrating by Parts Multiple Times
11. Integration by Parts and the Product Rule
12. Integrating ln(x)
13. Integration by Parts and Definite Integrals
2. The Return of the Identities
3. A Reciprocal Pythagorean Identity
4. Another Reciprocal Pythagorean Identity
5. The Original Pythagorean Identity
6. The Double Angle Identity for Sine
7. The Double Angle Identity for Cosine Part 1
8. The Double Angle Identity for Cosine Part 2
9. The Double Angle Identity for Cosine Part 3
10. The Double Angle Identity for Tangent
11. What Trig Functions Can We Now Integrate?
2. Finding the Area Between Two Curves
3. Finding the Area Between Two Points of Intersection
4. The Areas We Can’t Find Yet
5. What is a Trapezium?
6. Finding the Area of a Trapezium
7. Splitting an Area into Right Trapeziums
8. Finding the Width of the Strips
9. Finding the Boundary Points
10. Finding the Value of y at Each Boundary Point
11. Finding the Area Under the Curve
12. The Trapezium Rule Part I
13. The Trapezium Rule Part II
14. The Trapezium Rule Part III
15. The Trapezium Rule Part IV
16. Overestimating and Underestimating the Area
2. Convert to Cartesian Form then Integrate
3. Integrate Without Converting to Cartesian Form
4. Rewriting the Integral in Terms of x
5. Rewriting the Boundaries in Terms of t
6. Recapping the Strategies
2. Differential Equations in the Real World
3. Difficulties with Differential Equations
4. The Constant of Integration and Families of Curves
5. Methods for Differential Equations
6. Integrating as Normal
7. Solving Differential Equations
8. Why do we Use the Term ‘Solution’
9. The Differential Equations We Can Solve So Far
10. Recognising a Special Type of Differential Equation
11. Separation of Variables Part 1
12. Separation of Variables Part 2
13. Finding Particular Solutions to Differential Equations
14. Modelling with Differential Equations I
15. Modelling with Differential Equations II
16. Modelling with Differential Equations III
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]