# Evaluate the indefinite and definite integral of the function: (cubed root of x + 1/cubed root of x) dx May require integration by substitution

You need to evaluate the following indefinite integral such that:

int (root(3)(x+1))/(root(3)x) dx = int root(3)((x+1)/x) dx

You need to perform the following substitution such that:

(x+1)/x = t => ((x+1)'*x - (x+1)*x')/x^2 dx = dt

(x - x - 1)/x^2 dx = dt => -1/x^2 dx = dt

Changing the variable yields:

int - (root(3)t)/(1 - t)^2 dt

You need to perform the next substitution such that:

root(3) t = u => 1/(3root(3) t^2) dt = du => dt = 3u^2 du

Changing again the variable yields:

int - (u)/(1 - u^3)^2 (3u^2 du) = -3 int u^3/(1 - u^3)^2 du

You need to use partial fraction decomposition such that:

u^3/((1 - u)(1 + u + u^2))^2 = A/(1 - u) + B/(1 - u)^2 + (Cu + D)/(1 + u + u^2) + (Eu + F)/(1 + u + u^2)^2

u^3 = A(1-u)(1 + u + u^2)^2 + B(1 + u + u^2)^2 + (Cu + D)(1 - u)^2(1 + u + u^2) + (Cu + D)(1 - u)^2

u^3 = A(1 - u^3)(1 + u + u^2) + B(1 + u + u^2)^2 + (Cu + D)(1 - u)^2(1 + u + u^2) + (Cu + D)(1 - u)^2

Equating the coefficients of like parts yields:

A = B= (1/9)

C = -1/9

D = -1/3

E = F = 1/3

u^3/((1 - u)(1 + u + u^2))^2 = 1/(9(1 - u)) + 1/(9(1 - u))^2 + (-u - 3)/(9(1 + u + u^2)) + (u +1)/(3(1 + u + u^2))^2

Integrating both sides yields:

u^3/((1 - u)(1 + u + u^2))^2 = 1/(9(1 - u)) + 1/(9(1 - u))^2 + (-u - 3)/(9(1 + u + u^2)) + (u +1)/(3(1 + u + u^2))^2

int u^3/(1 - u^3)^2 du = int 1/(9(1 - u)) du+ int 1/(9(1 - u))^2 du+ int (-u - 3)/(9(1 + u + u^2)) du+ int (u +1)/(3(1 + u + u^2))^2 du

int u^3/(1 - u^3)^2 du = (1/9)ln|1-u|- (1/9)(1/(1 - u))+ int (-u - 3)/(9(1 + u + u^2)) du + int (u +1)/(3(1 + u + u^2))^2 du

int (-u - 3)/(9(1 + u + u^2)) du = -(1/9) int (u+1)/(u^2 + u + 1) du

Substituting v for u^2 + u +  1 yields:

dv = (2u+1)du

int (-u - 3)/(9(1 + u + u^2)) du = -(1/18) int dv/v = -1/18 ln|v|

int (-u - 3)/(9(1 + u + u^2)) du = -1/18 ln(u^2+u+1)+c

Substituting back root(3) t  for u and (x+1)/x  for t yields:

int root(3)((x+1)/x) dx = (arctan((2root(3)((x+1)/x) + 1)/sqrt3))/sqrt3 + x root(3)((x+1)/x) + (1/3) ln((root(3)((x+1)/x) + root(3)((x+1)/x)^2 + 1)/((root(3)((x+1)/x) - 1)^2)) + c

Hence, evaluating the given indefinite integral yields int root(3)((x+1)/x) dx = (arctan((2root(3)((x+1)/x) + 1)/sqrt3))/sqrt3 + x root(3)((x+1)/x) + (1/3) ln((root(3)((x+1)/x) + root(3)((x+1)/x)^2 + 1)/((root(3)((x+1)/x) - 1)^2)) + c.

Approved by eNotes Editorial Team