# Evaluate the indefinite integral integrate of (x^5+4x^3+3x^2-x+2)/(x^5+4x^3+4x)dx

sciencesolve | Teacher | (Level 3) Educator Emeritus

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You should add and subtract `4x`  to numerator such that:

`I = int (x^5 + 4x^3 + 4x - 4x + 3x^2 - x + 2)/(x^5+4x^3+4x)dx`

You should split the integral using its property of linearity such that:

`I = int (x^5 + 4x^3 + 4x)/(x^5 + 4x^3 + 4x) dx + int (3x^2 - 5x + 2)/(x^5 + 4x^3 + 4x)dx`

`I = int dx + int (3x^2 - 5x + 2)/(x^5 + 4x^3 + 4x)dx`

You need to write the factored form of numerator `3x^2 - 5x + 2`  such that:

`3x^2 - 5x + 2 = 0`

`x_(1,2) = (5 +- sqrt(25 - 24))/6`

`x_(1,2) = (5 +- 1)/6 => x_1 = 1 ; x_2 = 2/3`

`3x^2 - 5x + 2 = 3(x - 1)(x - 2/3)`

`int (3x^2 - 5x + 2)/(x^5 + 4x^3 + 4x)dx = int (3(x - 1)(x - 2/3))/(x^5 + 4x^3 + 4x)dx `

You need to factor out x to numerator such that:

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

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

You should use partial fraction decomposition such that:

`(3x^2 - 5x + 2)/(x(x^2+2)^2) = A/x + (Bx+C)/(x^2+2) + (Dx + E)/((x^2+2)^2))`

`3x^2 - 5x + 2 = A(x^2+2)^2 + (Bx+C)*(x^3+2x) + Dx^2 + Ex `

`3x^2 - 5x + 2 = Ax^4 + 4Ax^2 + 4A + Bx^4 + 2Bx^2 + Cx^3 + 2Cx + Dx^2 + Ex`

`3x^2 - 5x + 2 = x^4(A + B) + x^3(C) + x^2(4A + 2B + D) + x(2C + E) + 4A`

Equating the coefficients of like sides yields:

`A + B = 0`

`C= 0`

`4A + 2B + D = 3 => 4A - 2A + D = 3 => 2A + D = 3`

`2C + E = -5 => E = -5`

`4A = 2 => A = 1/2 => B = -1/2 => 2*(1/2) + D = 3=> D = 2`

`(3x^2 - 5x + 2)/(x(x^2+2)^2) = 1/(2x) - 1/(2(x^2+2)) + (2x-5)/((x^2+2)^2))`

Integrating both sides yields:

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

`int (3x^2 - 5x + 2)/(x(x^2+2)^2) dx = 1/2*ln|x| - (1/(2sqrt2))arctan(x/sqrt2) + int (2x-5)/((x^2+2)^2)) dx`

You should use the following substitution to solve `int (2x-5)/((x^2+2)^2)) dx`  such that:

`x^2+2 = t => 2xdx = dt`

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

`int (2x-5)/((x^2+2)^2)) dx = int (dt)/t^2- (5/8) ((2x)/(x^2+2) + sqrt2 arctan (x/sqrt2)) + c`

`int (2x-5)/((x^2+2)^2)) dx = -1/(x^2 + 2) - (5/8) ((2x)/(x^2+2) + sqrt2 arctan (x/sqrt2)) + c`

`int (3x^2 - 5x + 2)/(x(x^2+2)^2) dx = 1/2*ln|x| - (1/(2sqrt2))arctan(x/sqrt2) - 1/(x^2 + 2) - (5/8) ((2x)/(x^2+2) + sqrt2 arctan (x/sqrt2)) + c`

`I = x + 1/2*ln|x| - (1/(2sqrt2))arctan(x/sqrt2) - 1/(x^2 + 2) - (5/8) ((2x)/(x^2+2) + sqrt2 arctan (x/sqrt2)) + c`

Hence, evaluating the given integral yields `I = x + 1/2*ln|x| - (1/(2sqrt2))arctan(x/sqrt2) - 1/(x^2 + 2) - (5/8) ((2x)/(x^2+2) + sqrt2 arctan (x/sqrt2)) + c.`