Wednesday, March 23, 2022

March Calculus Madness Sweet Sixteen - Day 8: ∫ ln^2 x dx and ∫ ln^3 x dx

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Welcome to March Calculus Madness!


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Integration by parts to the rescue!


∫ ln^2 x dx


u = ln^2 x,   dv = dx

du = 2 * ln * 1/x dx,   v = x


∫ ln^2 x dx

= x * ln^2 x - ∫ 2 * ln x * 1/x * x dx

= x * ln^2 x - ∫ 2 * ln x dx


u = ln x, dv = 2 dx

du = 1/x dx, v = 2 * x


= x * ln^2 x - (2 * x * ln x - ∫ 1/x * 2 * x dx)

= x * ln^2 x - (2 * x * ln x - ∫ 2 dx)

= x * ln^2 x - 2 * x * ln x + ∫ 2 dx

= x * ln^2 x - 2 * x * ln x + 2 * x + C


∫ ln^3 x dx


u = ln^3 x dx, dv = dx

du = 3 * ln^2 x dx, v = x


= x * ln^3 x - ∫3 * ln^2 x * 1/x * x dx

= x * ln^3 x - 3 * ∫ ln^2 x dx

= x * ln^3 x - 3 * (x * ln^2 x - 2 * x * ln x + 2 * x + C)

(see the above)

=  x * ln^3 x - 3 * x * ln^2 x + 6 * x * ln x - 6 * x + D   

(where D = 3 * C, C and D are constants)


Eddie


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