# integrate 1/((7-6x)^1/2 - x^2)

sciencesolve | Teacher | (Level 3) Educator Emeritus

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You should use the substitution `sqrt(7-6x) = t ` such that:

`sqrt(7-6x) = t => -(6dx)/(2sqrt(7-6x)) = dt => -(3dx)/(sqrt(7-6x)) = dt => dx = -(tdt)/3`

`sqrt(7-6x) = t => 7 - 6x = t^2 => 7 - t^2 = 6x => x = (7 - t^2)/6`

`x^2 = ((7 - t^2)^2)/36`

You need to transform the integrand such that:

`1/(sqrt(7-6x) - x^2) = 1/(t- ((7 - t^2)^2)/36)`

`1/(t - ((7 - t^2)^2)/36) = 36/(36t - 49 + 14t^2 - t^4)`

`36t - 49 + 14t^2 - t^4= (t-1)(at^3+bt^2+ct+d)`

`36t - 49 + 14t^2 - t^4 = at^4+bt^3+ct^2+dt - at^3-bt^2-ct-d`

`-d = -49 => d = 49`

`a = -1`

`b - a = 0 => b = a => b = -1`

`c - b = 14 = > c + 1 = 14 => c = 13`

`at^3+bt^2+ct+d = -t^3 - t^2 + 13t +49`

`-t^3 - t^2 + 13t + 49 = -(t - 4.43)*(t^2 + 5.43t +11.06)`

You need to use partial fraction decomposition such that:

`36/(36t - 49 + 14t^2 - t^4) = A/(t-1) + B/(t - 4.43) + (Ct+D)/(t^2 + 5.43t +11.06)`

`36 = A(t - 4.43)(t^2 + 5.43t +11.06) + B(t^2 + 5.43t +11.06)(t-1) + (Ct+D)(t-1)(t - 4.43)`

`36 = At^3 + 5.43At^2 +11.06At - 4.43At^2 - 4.43*5.43At - 4.43*11.06A + Bt^3 + 5.43Bt^2 +11.06Bt - Bt^2 - 5.43Bt - 11.06B + (Ct+D)(t^2 - 4.43t - t + 4.43)`

`36 = At^3 + 5.43At^2 +11.06At - 4.43At^2 - 4.43*5.43At - 4.43*11.06A + Bt^3 + 5.43Bt^2 +11.06Bt - Bt^2 - 5.43Bt - 11.06B + Ct^3 - 4.43Ct^2 - Ct^2 + 4.43Ct + Dt^2 - 4.43Dt - Dt + 4.43D`

`A+B+C = 0 `

`5.43A - 4.43A + 5.43B - B - 4.43C - C + D = 0`

`11.06A- 4.43*5.43A + +11.06B - 5.43B + 4.43C - 4.43D - D = 0`

`- 4.43*11.06A - 11.06B + 4.43D = 36`

`int 36/(36t - 49 + 14t^2 - t^4) dt= int A/(t-1) dt+ int B/(t - 4.43)dt + int (Ct+D)/(t^2 + 5.43t +11.06)dt`

`int 36/(36t - 49 + 14t^2 - t^4) dt = Aln|t-1| + Bln|t - 4.43| +`  `ln|t^2 + 5.43t +11.06| + c`

Hence, evaluating the given integral yields:

`int 1/(sqrt(7-6x) - x^2) dx = Aln|sqrt(7-6x)-1| + Bln|sqrt(7-6x) - 4.43| + ln|(7-6x) + 5.43sqrt(7-6x) +11.06| + c`