# Compute the following indefinite integral. 1/(x-1)(x+1)^2

sciencesolve | Teacher | (Level 3) Educator Emeritus

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You need to use partial fraction decomposition to write the fraction in simpler terms such that:

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

You need to bring fractions to a common denominator such that:

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

Since denominators are the same, then the numerator are like.

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

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

Equating the coefficients of like powers yields:

`A+B=0 =gt A=-B`

`2A+C=0 =gt C=-2A`

`A-B-C=1 =gt A+A+2A=1 =gt A = 1/4 =gt B = -1/4, C = -1/2`

You need to calculate the integrals of simpler terms:

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

`int (dx)/((x-1)(x+1)^2) = 1/4(ln|(x-1)/(x+1)|) - int (dx)/(2(x+1)^2)`

You need to use substitution to solve `int (dx)/(2(x+1)^2)`  such that:

`x+1=y =gt dx=dy =gt int (dx)/(2(x+1)^2) = int (dy)/(2(y)^2) = -1/2y + c`

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

`int (dx)/((x-1)(x+1)^2) = 1/4(ln|(x-1)/(x+1)|) - 1/(2(x+1))+ c`

Hence, evaluating the indefinite integral yields `int (dx)/((x-1)(x+1)^2) = ln root(4)((x-1)/(x+1)) - 1/(2(x+1))+ c.`