Compute the antiderivatives of the following functions

(a) f(x) = x(x + 1)(x + 2)

(b) f(x) = (x^2 + x + 1) / √x

(c) f(x) = (√x + 1)(x - √x + 1)

(d) f(x) = [3√ x + (1/ x3√x) ]^2

Please show work, thanks!

∫ x(x + 1)(x + 2) dx

∫ (x^2 + x)(x + 2) dx

∫ (x^3 + 2x^2 + x^2 + 2x) dx

∫ (x^3 + 3x^2 + 2x) dx

(1/4) * x^4 + x^3 + x^2 + C

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∫ [ (x^2 + x + 1) / √x ] dx

∫ [ x^2/√x + x/√x + 1/√x ] dx

∫ x^(2 - 1/2) + x^(1 - 1/2) + x^(-1/2) dx

∫ x^(3/2) + x^(1/2) + x^(-1/2) dx

x^(3/2 + 1)/(3/2 + 1) + x^(1/2 + 1)/(1/2 + 1) + x^(-1/2 + 1)/(-1/2 + 1) + C

x^(5/2)/(5/2) + x^(3/2)/(3/2) + x^(1/2)/(1/2) + C

(2/5) * x^(5/2) + (2/3) * x^(3/2) + 2 * x^(1/2) + C

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∫ ( √(x) + 1 ) ( x - √(x) + 1 ) dx

∫ ( √(x)x - √(x)√(x) + √(x) + x - √(x) + 1) dx

∫ ( x^(1 + 1/2) - x + x + 1) dx

∫ ( x^(3/2) + 1) dx

x^(3/2 + 1)/(3/2 + 1) + x + C

x^(5/2)/(5/2) + x + C

(2/5) * x^(5/2) + x + C

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∫ [3√ x + (1/ (x^3 * √(x)) ]^2 dx <=== assuming u meant

∫ [3√ x + (1/ (x^(3 + 1/2) ) ]^2 dx

∫ [3√ x + (1/ (x^(7/2) ) ]^2 dx

∫ ( 3√(x) + x^(-7/2) )^2 dx

∫ ( 3√(x) + x^(-7/2) )( 3√(x) + x^(-7/2) ) dx

∫ ( 3√(x)3√(x) + 3√(x) * x^(-7/2) + 3√(x) * x^(-7/2) + x^(-7/2)x^(-7/2) ) dx

∫ ( 9x + 6 * x^(1/2 + -7/2) + x^(-7/2 + -7/2) ) dx

∫ ( 9x + 6 * x^(-6/2) + x^(-14/2) ) dx

∫ ( 9x + 6x^-3 + x^(-7) ) dx

(9/2) * x^2 + 6x^(-3+1)/(-3+1) + x^(-7+1)/(-7+1) + C

(9/2) * x^2 + 6x^(-2)/(-2) + x^(-6)/(-6) + C

(9/2) * x^2 - 3x^(-2) - (1/6) * x^(-6) + C

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please e-mail me if u have a question.

multiply out if needed

write the power of x of each term explicitly, ie. 1/√x = x^(-1/2).

then apply integral(x^m) = x^(m+1) / (m+1)