# How to solve the indefinite integral of y=square root x/ (square root x - 1) ?

*print*Print*list*Cite

### 1 Answer

### User Comments

To determine the indefinite integral of the function y, I'll suggest to replace the variable x by t^2.

x = t^2

We'll differentiate both sides:

dx = 2tdt

Int sqrtx dx/(sqrt x - 1) = Int [t/(t - 1)]*(2t dt)

Int 2t^2 dt/(t - 1) = 2Int t^2dt/(t - 1)

2Int (t^2 - 1 + 1)dt/(t - 1) = 2Int (t^2 - 1)dt/(t-1) + 2Int dt/(t-1)

We'll calculate the 1st term:

2Int (t^2 - 1)dt/(t-1)

We'll replace the difference of 2 squares by the product (t-1)(t+1).

2Int (t^2 - 1)dt/(t-1) = 2Int (t-1)(t+1)dt/(t-1)

We'll simplify and we'll get:

2Int (t-1)(t+1)dt/(t-1) = 2Int (t+1)dt = 2Int t dt + 2Int dt

2Int (t+1)dt = 2t^2/2 + 2t + C

2Int (t+1)dt = t^2 + 2t + C

2Int (t^2 - 1 + 1)dt/(t - 1) = t^2 + 2t + 2 ln|t - 1| + C

The indefinite integral of the given function is:

**Int sqrtx dx/(sqrt x - 1) = x + 2sqrt x+ 2 ln|sqrt x- 1| + C**