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...

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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.`