Question

What is the integration of (2x+1)/(2x+3) ?

Answer 1

`int(2x+1)/(2x+3)dx=x-ln|2x+3|`

Explanation:

`int(2x+1)/(2x+3)dx`

= `int(2x+3-2)/(2x+3)dx`

= `int(1-2/(2x+3))dx`

= `intdx-int2/(2x+3)dx`

Now let `t=2x+3`, then `dt=2dx`

and our integral becomes

`intdx-int1/tdt`

= `x-lnt`

= `x-ln|2x+3|`

Answer 2

`int (2x+1)/(2x+3)dx = x- ln abs(2x+3)+C`

Explanation:

Substitute:

`u=2x+3`

`du = 2dx`

Then:

`int (2x+1)/(2x+3)dx = 1/2 int (2x+3-2)/(2x+3) (2dx)`

`int (2x+1)/(2x+3)dx = 1/2 int (u-2)/u du`

and using the linearity of the integral:

`int (2x+1)/(2x+3)dx = 1/2 int du - int (du)/u`

`int (2x+1)/(2x+3)dx = u/2 - ln abs u+C`

and undoing the substitution:

`int (2x+1)/(2x+3)dx = (2x+3)/2 - ln abs(2x+3)+C`

and as we can absorb the constant in `C`:

`int (2x+1)/(2x+3)dx = x- ln abs(2x+3)+C`

Answer 3

`I=x-ln|2x+3|+c`

Explanation:

Here,

`I=int ((2x+1)/(2x+3))dx`

`=int(2x+3-2)/(2x+3)dx`

`=int(2x+3)/(2x+3)dx-int2/(2x+3)dx`

`=int1dx-int(d/(dx)(2x+3))/(2x+3)dx`

`=x-ln|2x+3|+c`