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Welcome to March Calculus Madness!
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d/dx x^n ∙ √(1 + x)
Here we can make use the multiplication rule:
d/dx f(x) ∙ g(x) = f(x) ∙ g'(x) + f'(x) ∙ g(x)
In this case:
f(x) = x^n
g(x) = √(1 + x) = (1 + x)^(1/2)
Then:
f'(x) = n ∙ x^(n-1)
g'(x) = 1/2 ∙ (1 + x)^(-1/2)
And:
d/dx x^n ∙ √(1 + x)
= x^n ∙ 1/2 ∙ (1 + x)^(-1/2) + n ∙ x^(n-1) ∙ (1 + x)^(1/2)
For indefinite integrals, I will do two specific cases of n.
∫ x ∙ √(1 + x) dx (n = 1)
Using integration by parts:
u = x, dv = (1 + x)^(1/2) dx
du = dx, v = 2/3 ∙ (1+x)^(3/2)
∫ x ∙ √(1 + x) dx
= 2/3 ∙ (1+x)^(3/2) ∙ x - ∫ (1 + x)^(1/2) dx
= 2/3 ∙ (1+x)^(3/2) ∙ x - 2/3 ∙ (1 + x)^(3/2) + C
= 2/3 ∙ (1 + x)^(3/2) ∙ (x - 1) + C
∫ x^2 ∙ √(1 + x) dx (n = 2)
Let z = (1 + x)^(1/2)
dz = 1/2 ∙ (1+ x)^(-1/2) dx
2 ∙ (1+x)^(1/2) dz = dx
2 ∙ z dz = dx
z^2 = 1 + x
z^2 - 1 = x
z^4 - 2 ∙ z^2 + 1 = x^2
∫ x^2 ∙ √(1 + x) dx
= ∫ (z^4 - 2 ∙ z^2 + 1) ∙ z ∙ 2 ∙ z dz
= ∫ (z^4 - 2 ∙ z^2 + 1) ∙ 2 ∙ z^2 dz
= ∫ 2 ∙ z^6 - 4 ∙ z^4 + 2 ∙ z^2 dz
= 2/7 ∙ z^7 - 4/5 ∙ z^5 + 2/3 ∙ z^3 + C
= 2/7 ∙ (1 + x)^(7/2) - 4/5 ∙ (1 + x)^(5/2) + 2/3 ∙ (1 + x)^(3/2) + C
(z = (1 + x)^(1/2))
Eddie
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