# Could somebody tell me how to complete this integral by using Partial Fractions?   Integral of 1/(x^2)sqrt((x^2)-5) dx

You should come up with the following substitution, such that:

x^2 - 5 = t^2 => 2xdx = 2tdt => dx = (tdt)/(sqrt(t^2+5))

sqrt(x^2 - 5) = sqrt t^2 = t

Changing the variable, yields:

int 1/((t^2+5)*t))*(tdt)/(sqrt(t^2+5))

You may use partial  fraction decomposition, such that:

1/(t(t^2 + 5)) = A/t + (Bt + C)/(t^2+5)

1 = At^2 + 5A + Bt^2 + Ct

Equating the coefficients of like powers yields:

{(A+B = 0),(C = 0),(5A = 1):} => A=-B=1/5

1/(t(t^2+5)) = 1/(5t) - 1/(5(t^2+5))

Integrating both sides yields:

int 1/((t^2+5)*t))*(tdt)/(sqrt(t^2+5)) = (1/5)int (dt)/(sqrt(t^2+5)) - (1/5)int tdt/((t^2+5)sqrt(t^2+5))

You should come up with the substitution t^2 + 5 = u , such that:

2t dt = du => tdt = (du)/2

int tdt/((t^2+5)sqrt(t^2+5)) = (1/2)int du/(usqrt u)

int tdt/((t^2+5)sqrt(t^2+5)) = (1/2)int (du)/(u*u^(1/2))

int tdt/((t^2+5)sqrt(t^2+5)) = (1/2)int (du)/(u^(1/2+1))

int tdt/((t^2+5)sqrt(t^2+5)) = (1/2)int u^(-3/2)du

int tdt/((t^2+5)sqrt(t^2+5)) = (1/2) u^(-3/2+1)/(-3/2+1)

int tdt/((t^2+5)sqrt(t^2+5)) = (1/2) u^(-1/2)/(-1/2) = -1/(sqrtu)

Substituting back t^2 + 5 for u yields:

int tdt/((t^2+5)sqrt(t^2+5)) = -1/(sqrt(t^2+5))

int 1/((t^2+5)*t))*(tdt)/(sqrt(t^2+5)) = (1/5)int (dt)/(sqrt(t^2+5)) + 1/(sqrt(t^2+5))

int 1/((t^2+5)*t))*(tdt)/(sqrt (t^2+5)) = (1/5)ln(t + sqrt(t^2 + 5)) + 1/(sqrt(t^2+5)) + c

Hence, evaluating the given integral yields I = (1/5)ln(sqrt(x^2 - 5) + x) + 1/x + c

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