(Method 2) Integral of 1/(x*sqrt(x^4-1)) (substitution)

[@IntegralsForYou]

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[@Melek_OK.]

😍👏👏 excelente!!!!! 📓✌️

[@IntegralsForYou]

@@Melek_OK. Ya me dirás si te gustan los diseños! Para cualquier sugerencia o comentario puedes enviarme un correo a [e-mail] 😉

[@Melek_OK.]

Madre mía, recuerdo este mismo ejercicio en mi examen y yo tipo 😵‍💫🥸🤣🤣🤣

[@IntegralsForYou]

jaja es que a primera vista impresiona! Yo la he resuelto de dos modos: [Method 1] - Integral of 1/(x*sqrt(x^4-1)) - [u=1/sqrt(x^4-1)] - kzbin.info/www/bejne/g6PKY5yMituDsLs [Method 2] - Integral of 1/(x*sqrt(x^4-1)) - [u=x^2 & u=1/cos(t)] - kzbin.info/www/bejne/l4q8n6yEi8-ffsk

[@--222---]

Hi Integrals I have been trying to solve this integral 13/x^3 - 125, I have been able to find the values for A,B,C however Im struggling to solve the integrals after that, specifically the one with the trinomal at the bottom. could you help me?

[@IntegralsForYou]

Of course! Here you have the solution! It is the whole solution, if you want it from exactly what you asked, you can start from the section starting with "Complete squares for x^2+5x+25": 13/(x^3 -125) = 13/(x-5)(x^2+5x+25) = A/(x-5) + (Bx+C)/(x^2+5x+25) = = A(x^2+5x+25)/(x-5)(x^2+5x+25) + (Bx+C)(x-5)/(x^2+5x+25)(x-5) = = (Ax^2+5Ax+25A)/(x^3 -125) + (Bx^2+Cx-5Bx-5C)(x-5)/(x^3 -125) = = (Ax^2 + 5Ax + 25A + Bx^2 + Cx - 5Bx - 5C)/(x^3 -125) = = ( (A+B)x^2 + (5A+C-5B)x + (25A-5C) )/(x^3 -125) ==> 13/(x^3 -125) = ( (A+B)x^2 + (5A+C-5B)x + (25A-5C) )/(x^3 -125) 13 = (A+B)x^2 + (5A+C-5B)x + (25A-5C) 0x^2 +0x + 13 = (A+B)x^2 + (5A+C-5B)x + (25A-5C) ==> 0 = A + B ==> B = -A 0 = 5A + C - 5B ==> 0 = 5A + C - 5(-A) ==> 0 = 5A + C + 5A ==> C = -10A 13 = 25A - 5C ==> 13 = 25A - 5(-10A) ==> 13 = 25A + 50A ==> 13 = 75A ==> A = 13/75 ==> A = 13/75 B = -A = -13/75 C = -10A = -10(13/75) = -130/75 ==> 13/(x^3 -125) = (13/75)/(x-5) + ((-13/75)x - 130/75)/(x^2+5x+25) = (1/75)*[13/(x-5) - (13x-130)/(x^2+5x+25)] Integral of 13/(x^3 -125) dx = = Integral of (1/75)*[13/(x-5) - (13x-130)/(x^2+5x+25)] dx = = (1/75)*[ 13*Integral of 1/(x-5) dx - Integral of (13x-130)/(x^2+5x+25) dx ] = Derivative of x^2+5x+25 = 2x+5. Then: 13x - 130 = (13/2)*(2x+20) = (13/2)*(2x+5+15) = (1/75)*[ 13*Integral of 1/(x-5) dx - Integral of (13/2)*(2x+5+15)/(x^2+5x+25) dx ] = = (1/75)*[ 13*Integral of 1/(x-5) dx - (13/2)*( Integral of (2x+5)/(x^2+5x+25) dx + 15*Integral of 1/(x^2+5x+25) dx ) ] = = (13/75)*Integral of 1/(x-5) dx - (13/150)*Integral of (2x+5)/(x^2+5x+25) dx - (13/10)*Integral of 1/(x^2+5x+25) dx = = (13/75)*ln|x-5| - (13/150)*ln|x^2+5x+25| - (13/10)*Integral of 1/(x^2+5x+25) dx = Complete squares for x^2+5x+25: (x + 5/2)^2 = x^2 + 5x + 25/4 (x + 5/2)^2 + 75/4 = x^2 + 5x + 25/4 + 75/4 = x^2 +5x + 100/4 = x^2 +5x + 25 Now write x^2+5x+25 as 1 + (something)^2: x^2 + 5x + 25 = = 75/4 + (x + 5/2)^2 = = (75/4)[1 + (4/75)(x + 5/2)^2 ] = = (75/4)[1 + ((2/sqrt(75))x + (2/sqrt(75))(5/2))^2 ] = = (75/4)[1 + ((2/sqrt(75))x + 5/sqrt(75))^2 ] = = (75/4)[1 + ( (2x+5)/sqrt(75) )^2 ] = (13/75)*ln|x-5| - (13/150)*ln|x^2+5x+25| - (13/10)*Integral of 1/(x^2+5x+25) dx = = (13/75)*ln|x-5| - (13/150)*ln|x^2+5x+25| - (13/10)*Integral of 1/(75/4)[1 + ( (2x+5)/sqrt(75) )^2 ] dx = = (13/75)*ln|x-5| - (13/150)*ln|x^2+5x+25| - (13/10)*(4/75)*Integral of 1/[1 + ( (2x+5)/sqrt(75) )^2 ] dx = Substitution: u = (2x+5)/sqrt(75) du = 2/sqrt(75) dx ==> sqrt(75)/2 du = dx = (13/75)*ln|x-5| - (13/150)*ln|x^2+5x+25| - (13/10)*(4/75)*Integral of 1/(1+u^2) sqrt(75)/2 du = = (13/75)*ln|x-5| - (13/150)*ln|x^2+5x+25| - (13/10)*(4/75)*(sqrt(75)/2)*Integral of 1/(1+u^2) du = = (13/75)*ln|x-5| - (13/150)*ln|x^2+5x+25| - (13/25sqrt(3))*Integral of 1/(1+u^2) du = = (13/75)*ln|x-5| - (13/150)*ln|x^2+5x+25| - (13/25sqrt(3))*arctan(u) = = (13/75)*ln|x-5| - (13/150)*ln|x^2+5x+25| - (13/25sqrt(3))*arctan((2x+5)/sqrt(75)) + C Hope it helped! ❤

[@paulogabriel1251]

That was really good. I did it straight, replacing the X squared with the secant. I follow your videos weekly and this one was really more challenging. Post some "challenge" videos for the more advanced ones too hahaha (sorry for my bad English)

[@IntegralsForYou]

Your English is good, Paulo! Thanks for your comment... well, your comments! I have seen you around all these years! In general I do integrals people ask me but if you want some difficult ones you will find them in this playlist: kzbin.info/aero/PLpfQkODxXi4_3E8mGTTnulkkvL2LBbOFb But something tells me that you already knew about this playlist... 🤣

[@gustavoromero7556]

Ñ​@@IntegralsForYou

[@IntegralsForYou]

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