Notice the fraction first. Having one function above another should make you suspicious that the integral may produce a logarithm. When you test this by finding the derivative of the denominator, you will find that it closely resembles the numerator. We adjust the constant (coefficient) and integrate. There is no need to use the absolute value sign inside the logarithm because x²+4 can never be negative if x is a real number.

You will notice that the function buried in the denominator, (x²+4), has a derivative of 2x, which is very similar to the numerator, x. This strongly suggests a chain rule pattern and that is exactly the pattern that I pursue in evaluating this integral. I.e. I try to create the structure n.f'(x).[f(x)]^(n-1) within the integral so that the integral will be [f(x)]^n.

This third integral in Jim Coroneos' 100 Integrals list requires the use of partial fractions.

Because the derivative of x²-4 is 2x, we need to isolate the 5x in the numerator. This means that we separate the integral into ∫5x/(x²-4).dx + ∫2/(x²-4).dx. The first integral produces the logarithmic function that we expect. The second integral can be resolved using partial fractions. This produces two more logarithmic functions. All three logarithmic functions can be combined, and the result simplified, using logarithmic laws.

Jim Coroneos' fourth integral is evaluated very quickly indeed! A quick inspection shows that it conforms very closely to the chain rule pattern. After a minor adjustment of coefficients, the integral is found in seconds.

This fifth problem from Jim Coroneos' 100 Integrals list is to evaluate ∫sinx.sec³x.dx

We first recognise that -sinx is the derivative of cosx, and that sec³x = cosˉ³x. This means that this integral will only need minor adjustments to conform to the chain rule pattern that we would write as n.f'(x).[f(x)]^(n-1)

Recognising this allows us to evaluate this integral very rapidly (once we have converted the sec³x to cosˉ³x).

We now address the sixth integral from Jim Coroneos' list of 100 Integrals. In this case it is ∫cos²(x/2).dx

This integral (of the cosine squared of a half angle) looks difficult to evaluate until you realise that we have two identities that mention cos²x. They are:

● sin²x + cos²x ≡ 1

● cos2x ≡ cos²x - sin²x

Combining the two gives us cosx ≡ 2cos²(x/2) - 1 and, therefore, cos²(x/2) ≡ (cosx + 1)/2, a much easier expression to integrate! Watch it all unfold in this video.

We now encounter the seventh integral in Jim Coroneos' list of 100 Integrals. The integral in question is ∫x.sinx.dx

Because it contains the product of two quite simple functions, we will use integration by parts. We place the more complicated function within the derivative, so ∫x.sinx.dx becomes -∫x.d(cosx). Integrating by parts gives us -x.cosx + ∫cosx.dx and the rest is easy!

Jim Coronoes' eighth integral also 'lends itself' to integration by parts.

You will notice that ∫x.sec²2x.dx contains just two functions ... a simple 'x' and a secant-squared. Recalling that sec²x is the derivative of tanx, we can therefore convert this integral into ½∫x.d(tan2x). The rest follows fairly easily (if you have been introduced to this kind of integration).

Note that this integral is already in the form ∫u.dv (and we know how to calculate the derivative of inverse tangents), so we use integration by parts to get ∫tanˉ¹2x.dx = x.tanˉ¹2x - ∫x.d(tanˉ¹2x) = x.tanˉ¹2x - ∫x.2/(4x² + 1).dx

The rest, provided you know your integrals, is relatively easy.

We are making progress! Since this video is the tenth of the series, it means we are about 10% of the way through Jim Coroneos' 100 Integrals List (some of the later ones are a bit longer, so we are actually not quite there!).

Jim's tenth problem was to evaluate ∫x³/(x² + 1).dx

Almost instantly, you will realise that we have a fraction, and that the two functions are both polynomials. Moments after that, you would notice that the order of the numerator is higher than that of the denominator. This means that our first action will be to divide the functions. There are a number of ways of doing this (long division, synthetic division, using ax + b and simultaneous equations, inspection, etc.), but the result will be ∫x³/(x² + 1).dx = ∫x.dx - ∫x/(x² + 1).dx The first integral is elemenary/trivial and produces x²/2. The second integral produces a logarithm, -½ln(x² + 1). [Note that we do not have to use the absolute value sign inside the logarithm since x² + 1 is always positive.]

We have reached integral number eleven ... and this one requires that we evaluate ∫x/[(x+4)(x+2)].dx

Because the numerator and denominator are both polynomials, and because the numerator has order one less than the denominator, we expect (at least) to get a logarithmic function. This is because the derivative of a polynomial is of order one less than the original function so, we are close to having the pattern f'(x)/f(x).

The fact that the denominator is already factorised is a help because we will separate the expression using partial fractions. Using whichever method you prefer, you will find that ∫x/[(x+4)(x+2)].dx = ∫2/(x+4).dx - ∫1/(x+2).dx This produces TWO logarithmic functions which we may then combine into one neat expression using our logarithmic laws.

We are starting to encounter some rather more difficult integrals already. The twelfth integral in Jim Coroneos' List is quite daunting to look at: ∫[(x+1)(x-1)]/[(x-2)(x-3)].dx

If you look more closely, however, you will notice that both the numerator and denominator are monic quadratic expressions. That is, the coefficients of their x² terms are both 1. We therefore start by performing a division. The denominator divides into the numerator once, and the remainder is a linear function, 5x - 7. In this way, our original integral is now separated into two integrals. The first one is trivial. The second integral (of (5x-7)/[(x-2)(x-3)]) is less trivial. Because the denominator is factorised, however, we are able to resolve it using partial fractions. The two expressions that result integrate to give two logarithmic expressions which we can combine using our logarithmic laws.

Because the numerator has degree one less than the denominator, we expect a logarithmic function to be part of the solution. Also, because the denominator cannot be factorised (within the real number system), we cannot use partial fractions. Therefore, we rearrange the numerator to create two separate expressions/integrals.

The numerator of the first one will be the derivative of the denominator ... and the second expression will contain the 'remainder.' The first integral produces the logarithm that we expected. The second integral, after completing a square, produces an inverse tangent function.

Still following the theme of Jim Coroneos' 100 Integrals, we now address the fourteenth integral, ∫x³/(2x-1).dx.

In this case, because the numerator is of a higher degree than the denominator, and both are polynomials, we first carry out a division. This produces a quadratic expression plus a constant (remainder) over 2x-1. The quadratic equation is easy to integrate. The fraction produces a logarithmic expression.

Upon examining this integral closely, we note that, if the numerator was simply '1,' we could complete the square in the denominator. This would give us an expression of the form 1/√(a²-x²) ... which would produce an inverse sine function.

The extra x in the numerator appears bothersome, however, until we realise that the derivative of 1-x-x² is a linear function (i.e. of the form ax+b) ... similar to the form of the numerator ... so this integral has the basic form of a chain rule. Therefore, we rearrange the expression to get ∫(1+x).(1-x-x²)^(-½).dx

This resolves into a chain rule AND another integral that produces an inverse sine function.

Jim Coroneos' sixteenth integral is ∫1/[x²√(1-x²)].dx

This expression does not fit any of the 'normal' patterns that we would look for ... chain rule with f'(x),f(x)^n, or an inverse sine pattern (which would require just √(1-x²) in the denominator). We therefore simplify the radical by using a trigonometric substitution. We would like a perfect square within the radical, so the identity that is of use to us is either sin²x = 1 - cos²x or cos²x = 1 - sin²x. We choose to substitute x = sinθ. After simplification, the resulting integral is ∫cosec²θ.dθ ... which would be written ∫csc²θ.dθ in the USA.

This is a lesser known integral in schools here in NSW, Australia. Not only is ∫sec²θ.dθ = tanθ + C, but ∫cosec²θ.dθ = -cotθ + C You can verify this by finding the derivative of -cotθ. Substituting x = sinθ in this result provides our answer in terms of x. This little integral took us on an interesting journey where we learned about trigonometric substitution and about ∫cosec²θ.dθ.

In evaluating Jim Coroneos' seventeenth integral ∫1/[x√(a²+x²)].dx, we again have an expression that does not fit the 'normal' patterns that we would look for ... chain rule with f'(x).f(x)^n, etc.

We simplify the radical by choosing an appropriate trigonometric substitution. Since a² + a²tan²θ = a²sec²θ, we substitute x = a.tanθ

After simplification, the resulting integral is (1/a)∫cosecθ.dθ ... which would be written (1/a)∫cscθ.dθ in the USA. By multiplying numerator and denominator by (cotθ + cosecθ) we uncover a logarithmic structure. [You might also have replaced ∫cosecθ.dθ with ∫sinθ/(1-cos²θ).dθ and substituted p = cosθ. I recommend that you evaluate the integral using this method also.]

This integral only differs from the previous integral by one sign. The expression within the radical is a²-x² instead of a²+x². Therefore, we again have an expression that does not fit the 'normal' patterns that we would look for ... chain rule with f'(x).f(x)^n, etc.

We simplify the radical by choosing an appropriate trigonometric substitution. Because of the "-" sign within the radical, we use the Pythagorean Identity a²cos²θ = a² - a²sin²θ, and we therefore substitute x = a.sinθ After simplification, the resulting integral is (1/a)∫(1/sinθ).dθ which could be written as (1/a)∫cosecθ.dθ ... or (1/a)∫cscθ.dθ in the USA.

When we encountered this integral in the last video, we multiplied the numerator and denominator by (cotθ + cosecθ) and uncovered a logarithmic structure. This time we will follow a different path. We replace our integral (1/a)∫(1/sinθ).dθ with (1/a)∫sinθ/(1-cos²θ).dθ and then make the substitution m = cosθ. The remainder of the video pursues the result of that process.

Please note: My solution of 1/a{ln[a-√(a²-x²)] - lnx} could also be written in a number of other forms, depending on the values of a and x. You might, for example, multiply numerator and denominator in the first logarithm by the conjugate, a+√(a²-x²) to obtain 1/a{ln[(a²-(a²-x²))/(a+√(a²-x²))] - lnx} and this would result in 1/a{ln(x²) - ln(a+√(a²-x²)) - lnx} = 1/a{2lnx - ln(a+√(a²-x²)) - lnx} which equals 1/a{lnx - ln(a+√(a²-x²))}.

This integral only differs from the previous integral only by the reversal of the difference between squares within the radical. In other words, the expression inside the radical is x²-a² instead of a²-x².

We simplify the radical by choosing an appropriate trigonometric substitution. Because of the "-" sign within the radical, I chose to use the Pythagorean Identity a²tan²θ = a²sec²θ - a², and therefore substituted x = a.secθ. [You could, equally well, have chosen the identity a²cot²θ = a²cosec²θ - a², and substituted x = a.cosecθ ... or x = a.cscθ in the USA!].

PLEASE NOTE that I glossed over a potential problem here! The square root of tan²θ is actually the absolute value of tanθ because the radical takes the positive value only. This means that you should be very careful of the domain in which you are integrating. If this was a definite integral, the limits of the integral in x, when converted by the substitution x = a.secθ ... or, rather, θ = cosˉ¹(a/x), would provide the limits in θ. You should then note how tanθ behaves in this θ domain. Since this is an indefinite integral, however, I should have discussed BOTH possibilities and noted the appropriate domains for each ... but I felt that the video was going to be too long anyway. I will leave this as an exercise for you to deal with. If you contact me privately, I would be happy to explain further or, perhaps, to make another (explanatory) video to append to this one!

After simplification, we are astonished! The resulting integral is a surprisingly elementary (1/a)∫dθ. It is integrated easily and then we substitute for θ to produce our answer in terms of x.

This integral provides a wonderful opportunity for me to share about a principle of substitution ... that, often, it is better to use the simpler substitution (here, u = √x instead of u = √x + 1).

The first nine minutes of the video is devoted to discussing the two substitution options that we have with this integral. If you wish to skip all this preamble and go straight to my solution, just start at 9:10.

I chose to use the simpler substitution, u = √x. The resulting integral then requires that we undertake a simple polynomial division. Once that step has been completed, the solution appears very rapidly!

I hope that you learn something useful from evaluating this integral and encourage you, as a home exercise, to evaluate the integral using the alternative substitution (u = √x + 1). You will learn some good mathematical lessons along the way!

I had the privilege of being one of Graeme’s students in 2005. As a creatively minded individual I struggled greatly with Mathematics and was falling behind in my HSC studies. Under his careful and creative tutelage I began to truly understand Mathematics for the first time in my life. Graeme patiently taught me to master mathematical concepts in ways I could understand. In the week intensive I spent with him he taught me the entire HSC course, and I went on to receive a Band 6. I cannot recommend him highly enough.