solve integral by means of substitute x explain step by stepintegral x^2/(x^2-6x+10)

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sciencesolve | Teacher | (Level 3) Educator Emeritus

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I suggest you to add to numerator `-6x + 10 ` such that:

`int (x^2dx)/(x^2-6x+10) = int ((x^2 -6x + 10 + 6x -10)dx)/(x^2-6x+10)`

You need to split the integrand such that:

`int (x^2dx)/(x^2-6x+10) = int ((x^2 -6x + 10)dx)/(x^2-6x+10) + int ((6x -10)dx)/(x^2-6x+10)`

`int (x^2dx)/(x^2-6x+10) = int dx + int ((6x -10)dx)/((x-3)^2+1)`

You should come up with the substitution`x-3=t =gt x = t+3 =gt dx=dt`

`int ((6x -10)dx)/((x-3)^2+1) = 2int ((3x -5)dx)/((x-3)^2+1)`

You need to write the integral using the variable t:

`2int ((3t + 9-5)dt)/(t^2+1) = 2int ((3t + 4)dt)/(t^2+1)`

`2int ((3t + 4)dt)/(t^2+1) =6int (tdt)/(t^2+1) + 8 int (dt)/(t^2+1)`

You need to come up with substitution `t^2 + 1 = y =gt 2tdt = dy`

`6int (tdt)/(t^2+1) = 3int (2tdt)/(t^2+1) = 3 int (dy)/y = 3ln |y| + c`

`` `2int ((3t + 4)dt)/(t^2+1) = 3 ln(t^2+1) + 8 arctan ((x-3)^2 + 1) + c`

Hence, evaluating the integral using substitution method yields `int (x^2dx)/(x^2-6x+10) = ln ((x-3)^2+1)^3 + 8 arctan ((x-3)^2 + 1) + c` .

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