# `int_11^23dx/((x+1)sqrt(2x+3))` solve this definite integral.

### 1 Answer | Add Yours

Let;

`2x+3 = t^2`

`2dx = 2tdt`

`dx = tdt`

`2x+3 = t^2`

`x = (t^2-3)/2`

`x+1 = (t^2-3)/2+1 = (t^2-1)/2`

When x = 11 then t = 5

When x = 23 then t = 7

`int_11^23dx/((x+1)(sqrt(2x+3)))`

`= int_5^7t/((t^2-1)/2(t))dt`

`= int_5^7(2/(t^2-1))dt`

Using partial fractions we can get;

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

`int_5^7(1/(t-1)-1/(t+1))dt`

`= [ln(t-1)-ln(t+1)]_5^7`

`= [ln((t-1)/(t+1))]_5^7`

`= ln(6/8)-ln(4/6)`

`= ln(6/8xx6/4)`

`= ln(36/32)`

`= ln(9/8)`

** so the answer is **`ln9/8`

*Note:*

*ln(A/B) = lnA-lnB*

**Sources:**