Integral of 5x^2/(x^2+4)(x^2+9) (partial fraction decomposition)

[@IntegralsForYou]

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[@nikolaputic6991]

Can i recommend trying to solve Integral of arcsin(e^x)/e^3x by dx?

[@IntegralsForYou]

Hi! It is a long one...! But here you have the steps to solve it: Integral of arcsin(e^x)/e^3x dx = = Integral of arcsin(e^x)/(e^x)^3 dx = Substitution: t = e^3x dt = 3e^x dx = 3t dx ==> dt/3t = dx = Integral of arcsin(t)/t^3 dt/3t = = (1/3)*Integral of arcsin(t)/t^4 dt = = (1/3)*Integral of arcsin(t)*(t^-4) dt = Parts: u = arcsin(t) ==> du = 1/sqrt(1-t^2) dt dv = (t^-4) dt ==> v = t^-3/-3 = -1/3t^3 = arcsin(t)*(-1/3t^3) - (1/3)*Integral of (-1/3t^3)*1/sqrt(1-t^2) dt = = -arcsin(t)/3t^3 + (1/9)*Integral of 1/(t^3)sqrt(1-t^2) dt = = -arcsin(t)/3t^3 + (1/9)*Integral of (1/t^3)*1/sqrt(1-t^2) dt = = -arcsin(t)/3t^3 + (1/9)*Integral of (1/t^4)*t/sqrt(1-t^2) dt = Substitution: s = sqrt(1-t^2) ==> s^2 = 1-t^2 ==> t^2 = 1-s^2 ==> t^4 = (1-s^2)^2 ds = -2t/2sqrt(1-t^2) dt = -t/sqrt(1-t^2) dt ==> -ds = t/sqrt(1-t^2) dt = -arcsin(t)/3t^3 + (1/9)*Integral of 1/(1-s^2)^2 (-ds) = = -arcsin(t)/3t^3 - (1/9)*Integral of 1/(1-s^2)^2 ds = Partial fraction decomposition: 1/(1-s^2)^2 = 1/((1-s)(1+s))^2 = 1/((1-s)^2)((1+s)^2) = A/(1-s) + B/(1-s)^2 + C/(1+s) + D/(1+s)^2 = = A(1-s)((1+s)^2)/((1-s)^2)((1+s)^2) + B(1+s)^2/((1-s)^2)((1+s)^2) + C((1-s)^2)(1+s)/((1-s)^2)((1+s)^2) + D(1-s)^2/((1-s)^2)((1+s)^2) = = A(1-s)(1+2s+s^2)/(1-s^2)^2 + B(1+2s+s^2)/(1-s^2)^2 + C(1-2s+s^2)(1+s)/(1-s^2)^2 + D(1-2s+s^2)/(1-s^2)^2 = = A(1+2s+s^2-s-2s^2-s^3)/(1-s^2)^2 + B(1+2s+s^2)/(1-s^2)^2 + C(1-2s+s^2+s-2s^2+s^3)/(1-s^2)^2 + D(1-2s+s^2)/(1-s^2)^2 = = A(1+s-s^2-s^3)/(1-s^2)^2 + B(1+2s+s^2)/(1-s^2)^2 + C(1-s-s^2+s^3)/(1-s^2)^2 + D(1-2s+s^2)/(1-s^2)^2 = = [ A + As - As^2 - As^3 + B + 2Bs + Bs^2 + C - Cs - Cs^2 + Cs^3 + D - 2Ds + Ds^2 ]/(1-s^2)^2 = = [ (-A+C)s^3 + (-A+B-C+D)s^2 + (A+2B-C-2D)s + (A+B+C+D) ]/(1-s^2)^2 ==> 0 = -A+C ==> A = C 0 = -A+B-C+D ==> 0 = -A+B-A+D ==> 0 = -2A+B+D 0 = A+2B-C-2D ==> 0 = A+2B-A-2D ==> 0 = 2B-2D ==> 0 = B-D 1 = A+B+C+D ==> 1 = A+B+A+D ==> 1 = 2A+B+D ==> A = C 0 = -2A+B+D ==> 0 = -2A+B+B ==> 0 = -2A+2B ==> 0 = -A+B ==> A = B 0 = B-D ==> D = B 1 = 2A+B+D ==> 1 = 2A+B+B ==> 1 = 2A+2B ==> A = C A = B D = B 1 = 2A+2B ==> 1 = 2B+2B ==> 1 = 4B ==> B = 1/4 ==> A = 1/4 A = 1/4 D = 1/4 B = 1/4 ==> 1/(1-s^2)^2 = A/(1-s) + B/(1-s)^2 + C/(1+s) + D/(1+s)^2 = (1/4)/(1-s) + (1/4)/(1-s)^2 + (1/4)/(1+s) + (1/4)/(1+s)^2 = -arcsin(t)/3t^3 - (1/9)*Integral of [ (1/4)/(1-s) + (1/4)/(1-s)^2 + (1/4)/(1+s) + (1/4)/(1+s)^2 ] ds = = -arcsin(t)/3t^3 - (1/9)*Integral of (1/4)*[ 1/(1-s) + 1/(1-s)^2 + 1/(1+s) + 1/(1+s)^2 ] ds = = -arcsin(t)/3t^3 - (1/9)*(1/4)*Integral of [ 1/(1-s) + 1/(1-s)^2 + 1/(1+s) + 1/(1+s)^2 ] ds = = -arcsin(t)/3t^3 - (1/36)*Integral of [ 1/(1-s) + 1/(1-s)^2 + 1/(1+s) + 1/(1+s)^2 ] ds = = -arcsin(t)/3t^3 - (1/36)*[ Integral of 1/(1-s) ds + Integral of 1/(1-s)^2 ds + Integral of 1/(1+s) ds + Integral of 1/(1+s)^2 ds ] = = -arcsin(t)/3t^3 - (1/36)*[ -Integral of -1/(1-s) ds - Integral of -(1-s)^(-2) ds + Integral of 1/(1+s) ds + Integral of (1+s)^(-2) ds ] = = -arcsin(t)/3t^3 - (1/36)*[ -ln|1-s| - (1-s)^(-3)/(-3) + ln|1+s| + (1+s)^(-3)/(-3) ] = = -arcsin(t)/3t^3 - (1/36)*[ -ln|1-s| + 1/3(1-s)^3 + ln|1+s| - 1/3(1+s)^3 ] = = -arcsin(t)/3t^3 - (1/36)*[ -ln|1-sqrt(1-t^2)| + 1/3(1-sqrt(1-t^2))^3 + ln|1+sqrt(1-t^2)| - 1/3(1+sqrt(1-t^2))^3 ] = = -arcsin(e^x)/3e^3x - (1/36)*[ -ln|1-sqrt(1-e^2x)| + 1/3(1-sqrt(1-e^2x))^3 + ln|1+sqrt(1-e^2x)| - 1/3(1+sqrt(1-e^2x))^3 ] + C Hope I didn't do any mistake! 🙃

[@TheBhagwaBoy]

integrate (x ^ 2 + x)/((e ^ x + x + 1) ^ 2) dx

[@IntegralsForYou]

Hi! I'm trying but I don't find a way to solve it... do you have any hint? 🤞

[@TheBhagwaBoy]

@@IntegralsForYou I have it's answer but not solution , answer might give you some hint C - ln(1 + (x + 1) * e ^ (- x)) - 1/(1 + (x + 1) * e ^ (- x))

[@IntegralsForYou]

@@TheBhagwaBoy Oh, my god...! Well, knowing the answer I can see that we must do a u=1+(x+1)e^-x substitution. Since then, we can see that u=(e^x+x+1)/e^x=(e^x+x+1)*e^-x so we need to multiply the e^x+x+1 on the denominator by e^-x and then rewrite the expression in order to perform the substitution. We have to take into account that du=-x*e^-x dx for that: Integral of (x^2+x)/(e^x+x+1)^2 dx = = Integral of (x^2+x)/[(e^x+x+1)*(e^-x/e^-x)]^2 dx = = Integral of (x^2+x)/[(1+(x+1)*e^-x)/e^-x]^2 dx = = Integral of (x^2+x)/[(1+(x+1)*e^-x)^2/(e^-x)^2] dx = = Integral of (x^2+x)/[(1+(x+1)*e^-x)^2/e^-2x] dx = = Integral of (x^2+x)(e^-2x)/(1+(x+1)*e^-x)^2 dx = = Integral of x(x+1)(e^-x)(e^-x)/(1+(x+1)*e^-x)^2 dx = = Integral of [(x+1)(e^-x)/(1+(x+1)*e^-x)^2] x*(e^-x) dx = Substitution: u = 1+(x+1)e^-x ==> u-1 = (x+1)e^-x du = ( 0-(x+1)e^-x + e^-x ) dx = ( 0 - xe^-x - e^-x + e^-x )dx = -x*e^-x dx ==> -du = x*e^-x dx = Integral of (u-1)/u^2 (-du) = = Integral of (1-u)/u^2 du = = Integral of (1/u^2 - u/u^2) du = = Integral of (1/u^2 - 1/u) du = = Integral of 1/u^2 du - Integral of 1/u du = = Integral of u^(-2) du - Integral of 1/u du = = u^(-1)/(-1) - ln|u| = = -1/u - ln|u| = = -1/(1+(x+1)e^-x) - ln|1+(x+1)e^-x| + C Which can also be written as: -1/(1+(x+1)e^-x) - ln|1+(x+1)e^-x| + C = = -[1/(1+(x+1)e^-x)]*[e^x/e^x] - ln| (1+(x+1)e^-x)*(e^x/e^x)| + C = = -e^x/(e^x+x+1) - ln|(e^x+x+1)/e^x| + C = = -e^x/(e^x+x+1) - [ ln|e^x+x+1| - ln|e^x| ] + C = = -e^x/(e^x+x+1) - ln|e^x+x+1| + x + C What a nice integral... you made my day ❤

[@MikeMagTech]

I am completely self taught, and when I was first learning calculus I slacked off when it came to partial fractions, which was a mistake. I have since gone back and learned it, but it is still a weak area for me. It is a good lesson about NOT slacking off!

[@IntegralsForYou]

Hi, Mike! If you want some integrals that are solved by partial fraction decomposition, here you have the playlist for it: kzbin.info/aero/PLpfQkODxXi4-9Ts0IMGxzI5ssWNBT9aXJ 💪

[@MikeMagTech]

​@@IntegralsForYouThank you ; )

[@IntegralsForYou]

@@MikeMagTech My pleasure! ❤

[@TomoeBv]

Integral ∫(10s^2 + 4) / (s(s+1)(s+2)^2) ds

[@IntegralsForYou]

Hi! Here you have the solution: Partial fraction decomposition for (10x^2+4)/(x(x+1)(x+2)^2): (10x^2+4)/(x(x+1)(x+2)^2) = A/x + B/(x+1) + C/(x+2) + D/(x+2)^2 = = A(x+1)(x+2)^2/(x(x+1)(x+2)^2) + Bx(x+2)^2/(x(x+1)(x+2)^2) + Cx(x+1)(x+2)/(x(x+1)(x+2)^2) + Dx(x+1)/(x(x+1)(x+2)^2) = = A(x+1)(x^2+4x+4)/(x(x+1)(x+2)^2) + Bx(x^2+4x+4)/(x(x+1)(x+2)^2) + C(x^2+x)(x+2)/(x(x+1)(x+2)^2) + D(x^2+x)/(x(x+1)(x+2)^2) = = A(x^3+4x^2+4x+x^2+4x+4)/(x(x+1)(x+2)^2) + B(x^3+4x^2+4x)/(x(x+1)(x+2)^2) + C(x^3+x^2+2x^2+2x)/(x(x+1)(x+2)^2) + D(x^2+x)/(x(x+1)(x+2)^2) = = A(x^3+5x^2+8x+4)/(x(x+1)(x+2)^2) + B(x^3+4x^2+4x)/(x(x+1)(x+2)^2) + C(x^3+3x^2+2x)/(x(x+1)(x+2)^2) + D(x^2+x)/(x(x+1)(x+2)^2) = = (Ax^3+5Ax^2+8Ax+4A)/(x(x+1)(x+2)^2) + (Bx^3+4Bx^2+4Bx)/(x(x+1)(x+2)^2) + (Cx^3+3Cx^2+2Cx)/(x(x+1)(x+2)^2) + (Dx^2+Dx)/(x(x+1)(x+2)^2) = = ( Ax^3 + 5Ax^2 + 8Ax + 4A + Bx^3 + 4Bx^2 + 4Bx + Cx^3 + 3Cx^2 + 2Cx + Dx^2 + Dx )/(x(x+1)(x+2)^2) = = [ (A+B+C)x^3 + (5A+4B+3C+D)x^2 + (8A+4B+2C+D)x + 4A ]/(x(x+1)(x+2)^2) ==> (10x^2+4)/(x(x+1)(x+2)^2) = [ (A+B+C)x^3 + (5A+4B+3C+D)x^2 + (8A+4B+2C+D)x + 4A ]/(x(x+1)(x+2)^2) ==> 10x^2+4 = (A+B+C)x^3 + (5A+4B+3C+D)x^2 + (8A+4B+2C+D)x + 4A ==> 0x^3 + 10x^2 + 0x + 4 = (A+B+C)x^3 + (5A+4B+3C+D)x^2 + (8A+4B+2C+D)x + 4A ==> 0 = A + B + C ==> C = -A - B ==> C = -1-B 10 = 5A + 4B + 3C + D 0 = 8A + 4B + 2C + D 4 = 4A ==> A = 1 ==> C = -1-B 10 = 5A + 4B + 3C + D ==> 10 = 5 + 4B + 3(-1-B) + D ==> 10 = 5 + 4B - 3 - 3B + D ==> 10 = 2 + B + D ==> 8 - B = D 0 = 8A + 4B + 2C + D ==> 0 = 8 + 4B + 2(-1-B) + (8-B) ==> 0 = 8 + 4B - 2 - 2B + 8 - B ==> 0 = 14 + B ==> B = -14 A = 1 ==> C = -1-B = -1-(-14) = -1+14 = 13 D = 8 - B = 8 - (-14) = 8 + 14 = 22 B = -14 A = 1 ==> A = 1 B = -14 C = 13 D = 22 ==> (10x^2+4)/(x(x+1)(x+2)^2) = A/x + B/(x+1) + C/(x+2) + D/(x+2)^2 = 1/x - 14/(x+1) + 13/(x+2) + 22/(x+2)^2 Integral of (10x^2+4)/(x(x+1)(x+2)^2) dx = = Integral of [ 1/x - 14/(x+1) + 13/(x+2) + 22/(x+2)^2 ] dx = = Integral of 1/x dx - 14*Integral of 1/(x+1) dx + 13*Integral of 1/(x+2) dx + 22*Integral of 1/(x+2)^2 dx = = Integral of 1/x dx - 14*Integral of 1/(x+1) dx + 13*Integral of 1/(x+2) dx + 22*Integral of (x+2)^(-2) dx = = ln|x| - 14*ln|x+1| + 13*ln|x+2| + 22*(x+2)^(-1)/(-1) = = ln|x| - 14*ln|x+1| + 13*ln|x+2| - 22/(x+2) + C ❤

[@TomoeBv]

@@IntegralsForYou thankssss !

[@IntegralsForYou]

@@TomoeBv My pleasure! ❤