# Evaluate the definite integral integrate from 1 to 3 of ((x^2-4x+3)/(x(x+1)^2)dx

sciencesolve | Teacher | (Level 3) Educator Emeritus

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You need to use partial fraction decomposition to simplify the evaluation of the given integral such that:

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

`x^2-4x+3 = A(x+1)^2 + Bx(x+1) + Cx`

`x^2-4x+3 = Ax^2 + 2Ax + A + Bx^2 + Bx + Cx`

`x^2-4x+3 = x^2(A+B) + x(2A+B+C) + A`

Equating the coefficients of like powers yields:

A+B = 1

`2A+B+C= -4 => 6 - 2 + C = -4 => C = -8`

`A = 3 => B = 1 - 3 => B = -2`

`(x^2-4x+3)/(x(x+1)^2) = 3/x- 2/(x+1)- 8/((x+1)^2)`

Integrating both sides yields:

`int (x^2-4x+3)/(x(x+1)^2)dx = 3int 1/x dx- 2int 1/(x+1) dx- 8int 1/((x+1)^2) dx`

`int (x^2-4x+3)/(x(x+1)^2)dx = 3 ln|x| - 2ln|x+1| - 8int 1/((x+1)^2) dx`

You should use the following substitution x+1 = t such that:

`x+1 = t => dx = dt`

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

`int t^(-2) dt = -1/t + c`

Substituting back x+1 for t yields:

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

`int (x^2-4x+3)/(x(x+1)^2)dx = 3 ln|x| - 2ln|x+1|+ 8/(x+1) + c`

Using logarithmic identities yields:

`int (x^2-4x+3)/(x(x+1)^2)dx = ln |x^3/((x+1)^2)| + 8/(x+1) + c`

You may evaluate the definite integral using the following formula such that:

`int_(x_1)^(x_2) f(x)dx = F(x_2)- F(x_1)`

Reasoning by analogy yields:

`int_1^3 (x^2-4x+3)/(x(x+1)^2)dx = ln |3^3/((3+1)^2)| + 8/(3+1) - ln |1^3/((1+1)^2)|- 8/(1+1)`

`int_1^3 (x^2-4x+3)/(x(x+1)^2)dx = ln (27/16) + 2 - ln(1/4) - 4`

`int_1^3 (x^2-4x+3)/(x(x+1)^2)dx = ln(27/16*4) - 2`

`int_1^3 (x^2-4x+3)/(x(x+1)^2)dx = ln(27/4) - 2`

Hence, evaluating the definite integral yields `int_1^3 (x^2-4x+3)/(x(x+1)^2)dx = ln(27/4) - 2.`