Evaluate the integral integrate of ((6x^2-2x-1)/(4x^3-x))dx
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You need to factor out x to denominator such that:
`4x^3-x = x(4x^2 - 1)`
Notice that you may convert the difference of squares `4x^2 - 1` into a product, using the formula `a^2 - b^2 = (a-b)(a+b)` such that:
`4x^2 - 1 = (2x-1)(2x+1)`
You need to use partial fraction decomposition such that:
`(6x^2-2x-1)/(4x^3-x) = (6x^2-2x-1)/(x(2x-1)(2x+1)) = A/x + B/(2x-1) + C/(2x+1)`
`6x^2-2x-1 = A(4x^2 - 1) + Bx(2x+1) + Cx(2x-1)`
`6x^2-2x-1 = 4Ax^2 - A + 2Bx^2 + Bx + 2Cx^2 - Cx`
`6x^2-2x-1 = x^2(4A + 2B + 2C) + x(B - C)- A`
Equating the coefficients of like powers yields:
`4A + 2B + 2C = 6 => 2A + B + C = 3`
`B - C = -2`
`A = 1 => 2 + B + C = 3 => B + C = 1`
`B - C + B + C = -2 + 1 => 2B = -1 => B = -1/2 => C = 1+1/2 = 3/2`
`(6x^2-2x-1)/(x(2x-1)(2x+1)) = 1/x- 1/(2(2x-1)) + 3/(2(2x+1))`
You need to integrate both sides such that:
`int (6x^2-2x-1)/(x(2x-1)(2x+1)) dx= int 1/x dx- int 1/(2(2x-1)) dx+ int 3/(2(2x+1)) dx`
`int (6x^2-2x-1)/(x(2x-1)(2x+1)) dx = ln |x| - (1/2) int (dx)/(2x-1) + (3/2) int (dx)/(2x+1)`
You should use the following substitution to solve `int (dx)/(2x-1)` such that:
`2x - 1 = t => 2dx = dt => dx = (dt)/2`
Changing the variable yields:
`int (dx)/(2x-1) =(1/2)int (dt)/t = 1/2 ln |t| + c`
Substituting back `2x - 1` for t yields:
`int (dx)/(2x-1) = 1/2 ln |2x-1| + c`
Reasoning by analogy yields:
`int (dx)/(2x+11) = 1/2 ln |2x+1| + c`
`int (6x^2-2x-1)/(x(2x-1)(2x+1)) dx = ln |x| - (1/4) ln |2x-1| + (3/4)ln |2x+1| + c`
Hence, evaluating the given indefinite integral yields `int (6x^2-2x-1)/(x(2x-1)(2x+1)) dx = ln |x| - (1/4) ln |2x-1| + (3/4)ln |2x+1| + c.`
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