Could somebody tell me how to complete this integral by using Partial Fractions?

 

Integral of 1/(x^2)sqrt((x^2)-5) dx

Expert Answers

An illustration of the letter 'A' in a speech bubbles

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`

See eNotes Ad-Free

Start your 48-hour free trial to get access to more than 30,000 additional guides and more than 350,000 Homework Help questions answered by our experts.

Get 48 Hours Free Access
Approved by eNotes Editorial