Integration by completing the square (hard)

blackpenredpen,
Integral battles!

[blackpenredpen]

If you have an integral battle idea, tweet me @blackpenredpen

[voteforno.]

Cute. If you use partial fractions on the first integral, you get arctan(2x + sqrt(3)) + arctan(2x - sqrt(3)) which leads to the fact that this function must differ from arctan(x - 1/x) by a constant. There is a subtlety though, as the latter is discontinuous at x = 0. By using a couple trig identities, one can discover that this constant is +/- pi/2, with the sign depending on whether x > 0 or < 0.

[emperorpingusmathchannel]

Putnam and beyond is so good. It has everything. Get the free pdf though.

[Kdd]

Who is better:
Blackpenredpen: LIKE
Dr Peyam:Comment
(No one is allowed to vote both
But still u can LOLL!! :))
Edit: I will also do the same voting process on dr. Peyam's channel.

[FaerieDragonZook]

Alternate way to do the first integral:
Denominator: x^4 -x^2+1 = (x^4+2x^2+1)-3x^2 = (x^2+1)^2 - 3x^2 = (x^2+1+sqrt3 x)(x^2+1-sqrt3 x)
Numerator: (x^2+1) = 1/2 (2x^2+2) = 1/2[(x^2+1+sqrt3 x) + (x^2+1-sqrt3 x)].
The integral is in the form 1/2 int((u+v)/(uv) dx), which easily breaks down by partial fractions into 1/2 int((1/u + 1/v)dx). Let y = 2x+sqrt3 and z = 2x-sqrt3. Then the integral turns out to be equivalent to int(dy/(y^2+1) + dz/(z^2+1)), with solution atan(2x+sqrt3) + atan(2x-sqrt3) + C.

[divyanshuraj]

Consider a triangle ABC.The perpendicular dropped from A onto BC is not shorter than BC, and the perpendicular dropped onto AC from B is not shorter than AC.Find angles of triangle ABC?


IIT maths quiz exam question....a challenge for bprp!!

[沈博智-xy]

I divided the numerator and denominator by sin^2(x) instead.
getting arctan(-x/sinx) + C
or -arctan(x/sinx) + C

[Prox-wbtk]

You are really awesome SIR!!I love the way you solve the problems.😀

[abdullahifarah]

The best way to exercise your algebra.

[seikomyazawa]

I really miss the time you had that big white board..

[ayushsaxena]

Thanks man understood the Qs
Now can solve my assignment on my own..

[MathemaTeach]

I heard him whisper indefinite integral.. hahaha 😂 what is his name again?

[ianluebbers]

Took me a bit to get it, but i was able to solve the second after watching the first!

[soumyanamdeo]

Diffrential Calculus is bit easy then Integrals beacuse in diffrential calculus you are free to use chain rule, no matter what type of composite function is .

[infinitygaming]

The title gave me a hint to solve them

[electrovector]

When you see Tibees at Patreons😅. You realize that you are a community

[kingbeauregard]

The integral on the left: is there a particular strategy you used to find that, or is this more of a convenient situation that can't really be expanded into a strategy?

[Rahulsingh-lwhk]

Blackpenredpen loves integration just like anything ...

[hc_]

On the left integral, I multiplied both top and bottom by 1+x^2, manipulating the top I got x^4 - x^2 + 1 + 3x^2, and after splitting I got arctan(x^3) + arctan(x) + C

[気分によって面積が変わる-bj]

Wow! How do you come up with such a great replacement?