Jim Coroneos' 100 Integrals ~ 027 ~ ∫1/(1+x²)².dx

Partly to honour Jim, and partly to fulfil an international need, I have decided to produce 100 videos, showing how to solve his 100 integration 'problems.' I hope you find the videos useful!

This twenty-seventh problem is to evaluate ∫1/(1+x²)².dx

Since we have a denominator that contains the sum of squares (1+x²), our first step in evaluating this integral will be to use the trigonometric substitution, x = tanθ. Simplifying this gives us ∫cos²θ.dθ.

The next step is to convert this trigonometric square expression into a double angle form ∫(cos2θ + 1)/2.dθ. We then evaluate the integral and obtain, in its simplest form, (sinθ.cosθ + θ)/2 + C.

Using our original substitution x = tanθ to evaluate sinθ and cosθ allows us to write our solution in terms of x ...

∫1/(1+x²)².dx = x/2(1 + x²) + (tan‾¹x)/2 + C OR
∫1/(1+x²)².dx = x/2(1 + x²) + (arctanx)/2 + C

~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~

If you wish to be kept up to date with what I am producing on the website (ad free, spam free, cost free mathematics and study materials), please add your name to the mailing list there.

Best wishes for your study and your mathematics!

Thank you.

[holyshit]

I would recommend two approaches
1. Rewrite numerator as 1=1+x^2 - x^2
You will get two integrals in one of them numerator will cancel with denominator the other one can be calculated by parts
2. Use isolation of rational part of integral
= \frac{P_{1}(x)}{Q_{1}(x)} +
Now we know that Q(x) = Q_{1}(x)Q_{2}(x) and Q_{1}(x) = GCD(Q(x), Q'(x))
You can assume that degree of numerators are less than corresponding denominators and use undetermined coefficients for numerators P_{1}(x) and P_{2}(x)
Once you take derivative of both sides of equation above you can calculate numerators
\frac{P(x)}{Q(x)} = + \frac{P_{2}(x)}{Q_{2}(x)}
You can simplfy it further introducing new polynomial
H(x) =
P(x) = P_{1}'(x)Q_{2}(x) - P_{1}(x)H(x) + P_{2}(x)Q_{1}(x)

[PLCO]

Pasé días buscando la manera con la que se resolvía esta Integral, muchas gracias por el video. Ya tienes un nuevo suscriptor excelente explicación!.

[MrVoayer]

Thinking about all the trigonometric identities and choosing the proper one could indeed be challenging. But, with practice comes the knowledge ! Obviously, we had better know all those identities by heart. And this is not the first integral in the "Jim Coroneos' 100 integrals" series that kindly suggests it. Coming back from angle Theta to x variable may be a challenge, too.Thank you for patient, step by step explanation, dear Graeme.

[holyshit]

Not so long time ago i played with following things
I found in my tables following integral
integral of on interval [theta, pi]
If you want to play with it I suggest to derive recurrence relation using trigonometric idenitites like
sin(A+B) = sin(A)cos(B)+cos(A)sin(B)
then use integration by parts
and dont forget base cases
Once you have recurrence relation you can solve it with ordinary generating function
I calculated this integral in this way but maybe you find other one
When I played with orthogonalization and i got integral of x^n/sqrt(1-x^2) on interval [-1;1]
You could use x=cos(theta) substitution but in my opinion it is better to derive recurrence relation by parts and solve it

Do not trust in Wolfram Alpha
For example I have integral of x/sqrt((x+2)^2+exp(x)) dx
it evaluates to x - 2ln((x+2)^2+exp(x))+C
but Wolfram Alpha claimes that antiderivative cannot be expressed in terms of known functions
I tried to find coefficients of Chebyshov polynomial
I solved recurrence relation with exponential generating function and i got sum
sum k from m to [n/2] C[n, 2k]*C[k, m]
and Wolfram Alpha give incorrect result at least for n=0
(Yes I know derivation with complex numbers would be shorter but i tried to avoid them)

[carlosalexandre]

Thanks, I'm Brazilian but i understand and learn.

[harishd]

My first idea when I saw this integral was that the denominator is the product of two polynomials so I thought of partial fractions.. Could we also do this by partial fractions?