Integrating 1/-(x-2) and 1/(x-2) - What To Do?

[cmab]

Ok, I have to integrate this

[int a=0 b=3] 1 / sqrt[abs(x-2)] dx

what should i do ? do the improper integral of 1/-(x-2) and 1/(x-2) ?

 

[Zurtex]

Do you know how to work this out:

\int \frac{1}{\sqrt{x-2}}dx

If so just split the function you have into the 2 different functions it is composed of and intergrate over the relevant areas.

 

[thepatientmental]

To integrate 1/-(x-2) and 1/(x-2), you can use the substitution method. Let u = x-2, then du = dx. Substituting this into the integral, we get:

[int a=0 b=3] 1 / sqrt[abs(x-2)] dx = [int a=-2 b=1] 1 / sqrt[abs(u)] du

Now, we can split this into two integrals:

[int a=-2 b=1] 1 / sqrt[abs(u)] du = [int a=-2 b=0] 1 / sqrt[-u] du + [int a=0 b=1] 1 / sqrt du

Using the substitution method again, we can let v = -u for the first integral and v = u for the second integral. This gives us:

[int a=0 b=3] 1 / sqrt[abs(x-2)] dx = [int a=0 b=2] 1 / sqrt[v] dv + [int a=0 b=1] 1 / sqrt du

The first integral can be evaluated as 2*sqrt[v] from 0 to 2, which gives us 2*sqrt[2]. The second integral can be evaluated as 2*sqrt from 0 to 1, which gives us 2*sqrt[1] - 2*sqrt[0] = 2.

Therefore, the final result is 2*sqrt[2] + 2.