Integration — Page 1 — Tinkutara Math Forum

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∫_0 ^1 ((ln (x+1))/(x^2 +1))dx=?

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Question 228881

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∫e^x^2 .dx

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Question 228655

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∫((x^6 ln^2 (x))/((e^(2 ) −1)( )) dx (5/(96)) ((ζ(3))/ ) +(1/ ) ((ζ(5))/ ) −((77)/(3840)) ((ζ(7))/ ) −((11)/(1920)) (1/ ) −((47)/ ) (1/ ) Anyone pls prove with full understandable solution. Thank u

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Just want to share my work here. 1. ∫ e^x sin(ax) dx Let I = ∫ e^x sin(ax) dx =^(IBP) e^x sin(ax) − a∫ e^x cos(ax) dx =^(IBP) e^x sin(ax) − a[e^x cos(ax) + a∫ e^x sin(ax) dx] = e^x sin(ax) − ae^x cos(ax) − a^2 I I = e^x sin(ax) − ae^x cos(ax) − a^2 I (a^2 + 1)I = e^x sin(ax) − ae^x cos(ax) I = ((e^x sin(ax))/(a^2 + 1)) − ((ae^x cos(ax))/(a^2 + 1)) + C determinant (((∫ e^x sin(ax) dx = ((e^x sin(ax))/(a^2 + 1)) − ((ae^x cos(ax))/(a^2 + 1)) + C))) 2. ∫ e^x cos(ax) dx Let I = ∫ e^x cos(ax) dx =^(IBP) e^x cos(ax) + a∫ e^x sin(ax) dx =^(IBP) e^x cos(ax) + a[e^x sin(ax) − a∫ e^x cos(ax) dx] = e^x cos(ax) + ae^x sin(ax) − a^2 I I = e^x cos(ax) + ae^x sin(ax) − a^2 I (a^2 + 1)I = e^x cos(ax) + ae^x sin(ax) I = ((e^x cos(ax))/(a^2 + 1)) + ((ae^x sin(ax))/(a^2 + 1)) + C determinant (((∫ e^x cos(ax) dx = ((e^x cos(ax))/(a^2 + 1)) + ((ae^x sin(ax))/(a^2 + 1)) + C))) Note: C is an arbitrary constant Open for corrections.

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I = ∫_1 ^2 ((2x^2 )/( (√((2x−1)(2x+2))))) dx = 2∫_1 ^2 (x^2 /( (√((2x−1)(2x+2))))) dx =^(t = 2x) (1/4)∫_2 ^( 4) (t^2 /( (√((t−1)(t+2))))) dt Let ξ(t−1) = (√((t−1)(t+2))) ⇒ t = ((2+ξ^2 )/(ξ^2 −1)) ⇒ dt = −((6ξ)/((ξ^2 −1)^2 )) dξ I = (1/4)∫_2 ^( (√2)) (((((2+ξ^2 )/(ξ^2 −1)))^2 )/(ξ(((2+ξ^2 )/(ξ^2 −1))−1)))(−((6ξ)/((ξ^2 −1)^2 ))) dξ I = (1/4)∫_(√2) ^( 2) ((((2+ξ^2 )^2 )/((ξ^2 −1)^2 ))/(3/(ξ^2 −1)))((6/((ξ^2 −1)^2 ))) dξ I = (1/2)∫_(√2) ^( 2) (((2+ξ^2 )^2 )/((ξ^2 −1)^3 )) dξ Q(x) = (ξ^2 −1)^3 ⇒ Q′(x) = 6ξ(ξ^2 −1)^2 ⇒ Q_1 (x) = (ξ^2 −1)^2 & Q_2 (x) = ξ^2 −1 ∫ (((2+ξ^2 )^2 )/((ξ^2 −1)^3 )) dξ = ((Aξ^3 +Bξ^2 +Cξ+D)/((ξ^2 −1)^2 )) + ∫ ((Eξ+F)/(ξ^2 −1)) dξ ((4+4ξ^2 +ξ^4 )/((ξ^2 −1)^3 )) = (((ξ^2 −1)(3Aξ^2 +2Bξ+C)−4ξ(Aξ^3 +Bξ^2 +Cξ+D))/((ξ^2 −1)^3 )) + ((Eξ+F)/(ξ^2 −1)) ⇒ A = (3/8), B = D = E = 0, C = −((21)/8), F = ((11)/8) I = (1/2)[((3ξ^3 −21ξ)/(8(ξ^2 −1)^2 ))∣_(√2) ^2 + ((11)/8)∫_(√2) ^( 2) (1/(ξ^2 −1))∣ dξ] = (1/2)[((3ξ^3 −21ξ)/(8(ξ^2 −1)^2 ))∣_(√2) ^2 + ((11)/8)((1/2)ln(∣((ξ−1)/(ξ+1))∣))∣_(√2) ^2 ] = (1/2)[−(1/4) + ((15(√2))/8) + ((11)/8)(((−ln(3)+ln(3−(√2)))/2))] = (1/2)[−(1/4) + ((15(√2))/8) + ((11)/(16))ln(((3−(√2))/3))] = ((11)/(32))ln(((3−(√2))/3)) +((15(√2))/(16)) −(1/8) determinant (((∫_1 ^2 ((2x^2 )/( (√((2x−1)(2x+2))))) dx = ((11)/(32))ln(((3−(√2))/3)) + ((15(√2))/(16)) − (1/8)))) Forgot it′s Question No. I kind of missed the past times...

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I = ∫ (1/(x^5 +1)) dx = ∫ (1/((x+1)(x^4 −x^3 +x^2 −x+1))) dx = ∫ [(1/(5(x+1))) − ((x^3 −2x^2 +3x−4)/(5(x^4 −x^3 +x^2 −x+1)))] dx = (1/5)∫ (1/(x+1)) dx − (1/5)∫ ((x^3 −2x^2 +3x−4)/(x^4 −x^3 +x^2 −x+1)) dx = (1/5)ln ∣x+1∣ + C_1 − (1/5)∫ ((x^3 −2x^2 +3x−4)/(x^4 −x^3 +x^2 −x+1)) dx I_2 = ∫ ((x^3 −2x^2 +3x−4)/(x^4 −x^3 +x^2 −x+1)) dx = ∫ ((x^3 −2x^2 +3x−4)/((x^2 +(ϕ−1)x+1)(x^2 −ϕx+1))) dx = ∫ [(((1−ϕ)x−2)/(x^2 +(ϕ−1)x+1)) + ((ϕx−2)/(x^2 −ϕx+1))] dx = ∫ [(((1−ϕ)x−2)/(x^2 +(ϕ−1)x+1))] dx + ∫ [((ϕx−2)/(x^2 −ϕx+1))] dx = ((1−ϕ)/2)ln ∣x^2 +(ϕ−1)x+1∣ − (√(2−ϕ))tan^(−1) (((2x+ϕ−1)/( (√(2+ϕ))))) − (ϕ/2)ln ∣x^2 −ϕx+1∣ − (√(3−ϕ))tan^(−1) (((2x−ϕ)/( (√(3−ϕ))))) + C_2 I = determinant ((((1/5)ln ∣x+1∣ − ((1−ϕ)/(10))ln ∣x^2 +(ϕ−1)x+1∣ + (√(2−ϕ))tan^(−1) (((2x+ϕ−1)/( (√(2+ϕ))))) + (ϕ/(10))ln ∣x^2 −ϕx+1∣ + ((√(3−ϕ))/5)tan^(−1) (((2x−ϕ)/( (√(3−ϕ))))) + C))) Where ϕ = Golden Ratio = (((√5) + 1)/2) I wonder if I made any mistakes here. Can someone check this?

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Question 227988

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Question 227871

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Help please I = ∫_0 ^( π) cos(x + cos(x)) dx

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Question 227271

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Question 227255

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Question 227184

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Question 227149

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Question 227146

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Question 227128

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prove: ∫_0 ^1 ((lnx)/(1+x^6 ))dx=(π^2 /(12(√3)))−(5/9)G

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∫_0 ^1 ⌊log_2 (x−2^(⌊log_2 x⌋) )⌋dx

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∫_0 ^π ((sin 2x)/3)(√(cos x+1)) dx =?

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A parabolic refector is formed by revolving the arc of the parabala y^2 =4ax from x=0 to x=h about the axis.If the diameter of the reflector is 2l.Show that the area of the reflecting surface is ((πl)/(6h^2 )){(l^2 +4h^2 )^(3/2) −l^3 }

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Question 226995

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Question 226994

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If I_n =∫(x^2 +a^2 )^n dx Show that I_n =(1/(2n+1))x(x^2 +a^2 )^n +2na^2 I_(n−1)

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Question 226952

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Evaluate ∫_0 ^∞ (dx/(1+x^2 ))

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Evaluate ∫((x^2 +2x−1)/(2x^3 +3x^2 −2x))dx

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∫_0 ^1 ((ln(1+x^2 ))/(1+x)) dx = ?

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By using concept of complex number show that tan 5θ=((tan^5 θ−10tan^3 θ+5tan θ)/(5tan^4 θ−10tan^2 θ+1))

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Approximate ∫_0 ^1 xe^x^2 dx with 6 ordinates. Use both rules Simpsons and Trapozoidal rules,hence evaluate and calculate the percentage error commetted for each case.Give comments