# Evaluate the indefinite integral integrate of (dx/(x^3+1))

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You should use the following formula to write the sum of cubes such that:

`a^3 + b^3 = (a+b)(a^2-ab+b^2)`

Resoning by analogy yields:

`x^3 + 1 = (x+1)(x^2-x+1)`

You need to use the partial fraction decomposition to make easier the evaluation of the integral.

`1/(x^3 + 1) = 1/(x+1)(x^2-x+1)`

`1/(x+1)(x^2-x+1) = A/(x+1) + (Bx + C)/(x^2-x+1)`

`1 = Ax^2 - Ax + A + Bx^2 + Bx + Cx + C`

`1 = x^2(A+B) + x(-A + B + C) + A + C`

Equating the coefficients of like powers yields:

`A+B = 0 => A=-B`

`-A + B + C = 0 => -2A + C = 0`

`A + C = 1 => 2A + 2C = 2`

`2A + 2C -2A + C = 2 => 3C = 2 => C = 2/3 => A = C/2 => A = 1/3 => B = -1/3`

`1/(x+1)(x^2-x+1) = 1/(3(x+1)) + (-x + 2)/(3(x^2-x+1))`

Integrating both sides yields:

`int 1/(x+1)(x^2-x+1) dx= int 1/(3(x+1)) dx+ int (-x + 2)/(3(x^2-x+1)) dx`

`int 1/(x+1)(x^2-x+1) dx = (1/3) ln|x+1| - (1/3)int x/(x^2-x+1) dx -(2/3)int dx/(x^2-x+1)`

You need to solve the integral `(1/3)int x/(x^2-x+1) dx` using substitution such that:

`(1/3)int x/(x^2-x+1) dx = (1/6)int (2x - 1 + 1)/(x^2-x+1) dx`

`(1/3)int x/(x^2-x+1) dx = (1/6)int (2x - 1)(x^2-x+1) dx + (1/6)int 1/(x^2-x+1) dx`

`x^2-x+1 = t => (2x - 1)dx = dt`

Changing the variable yields:

`(1/3)int x/(x^2-x+1) dx = (1/6)int dt/t +(1/6)int 1/(x^2-x+1) dx`

`(1/3)int x/(x^2-x+1) dx = (1/6) ln(x^2-x+1) + (1/6)int 1/(x^2-x+1) dx`

`int 1/(x+1)(x^2-x+1) dx = (1/3) ln|x+1| - (1/6) ln(x^2-x+1) - (1/6)int 1/(x^2-x+1) dx - (2/3)int dx/(x^2-x+1)`

`int 1/(x+1)(x^2-x+1) dx = (1/3) ln|x+1| - (1/6) ln(x^2-x+1) - (5/6)int 1/(x^2-x+1) dx`

You need to complete the square `x^2-x` such that:

`x^2-x = x^2 - 2*(1/2)*x + 1/4 - 1/4`

`x^2-x = (x - 1/2)^2 - 1/4`

`x^2-x+1 = (x - 1/2)^2 - 1/4 + 1`

`x^2-x+1 = (x - 1/2)^2 + 3/4`

`x^2-x+1 = (x - 1/2)^2 + sqrt3/2`

`(5/6)int 1/(x^2-x+1) dx = (5/6)int 1/((x - 1/2)^2 + sqrt3/2) dx`

You should use the following substitution such that:

`x - 1/2 = u => dx = du`

`(5/6)int 1/(x^2-x+1) dx = (5/6)int du/(u^2 + sqrt3/2)`

`(5/6)int 1/(x^2-x+1) dx = (5sqrt3/9) arctan 2sqrt3*x/3 + c`

`int 1/(x+1)(x^2-x+1) dx = (1/3) ln|x+1| - (1/6) ln(x^2-x+1) - (5sqrt3/9) arctan 2sqrt3*x/3 + c`

**Hence, evaluating the given indefinite integral yields `int 1/(x^3 + 1) dx = (1/3) ln|x+1| - (1/6) ln(x^2-x+1) - (5sqrt3/9) arctan 2sqrt3*x/3 + c.` **