We have to find the integral of 1/(x-1) dx + 1/(x-1)^2 dx - 2/(2x+3) dx.
Int [ 1/(x-1) dx + 1/(x-1)^2 dx - 2/(2x+3) dx ]
=> Int [ 1/(x-1) dx] + Int [1/(x-1)^2 dx] - Int [2/(2x+3) dx]
- Int [ 1/(x-1) dx]
let u = x - 1 ,...
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We have to find the integral of 1/(x-1) dx + 1/(x-1)^2 dx - 2/(2x+3) dx.
Int [ 1/(x-1) dx + 1/(x-1)^2 dx - 2/(2x+3) dx ]
=> Int [ 1/(x-1) dx] + Int [1/(x-1)^2 dx] - Int [2/(2x+3) dx]
- Int [ 1/(x-1) dx]
let u = x - 1 , du = dx
=> Int [ 1/u du ]
=> ln u
=> ln( x +1)
- Int [1/(x-1)^2 dx]
let u = x - 1, du = dx
=> Int[ 1/u^2 du]
=> -1/u
=> -1/(x - 1)
- Int [2/(2x+3) dx]
let u = 2x + 3 , du = 2*dx
=> Int [ 1/u du]
=> ln u
=> ln (2x + 3)
Therefore we get the required integral as ln( x +1) - 1/(x - 1) - ln (2x + 3) + C