# why can't "U Substitution" solve the following Integral? `int` -2 / (2x +1) Instead, solution is: -2 ln abs(2x + 1) Missing something???

### 2 Answers | Add Yours

Solve `int -2/(2x+1)dx` :

If you try u-substitution you have u=2x+1,du=2dx and the integral becomes `int -u^(-1)du` .

It looks like you could use the power rule `int u^n=u^(n+1)/(n+1)` , but there is a problem. The power rule rule works for all `n!= -1` . If n=-1 then you will be dividing by zero. ````

Thus for n=-1, `int u^(-1)du=ln|u|+C`

-----------------------------------------------------------------

Why a logarithm? Find the area under the curve `y=1/x` from 1 to 2, 1 to 3, and 1 to 6 using Riemann sums.

`int_1^2 1/x dx ~~.69314718`

`int_1^3 1/x dx ~~1.0986123`

`int_1^6 1/x dx ~~1.7917595`

So if `f(a)=int_1^a 1/x dx` then f(2)+f(3)=f(6). Try some other numbers, always using Riemann sums, and you will find that f(a*b)=f(a)+f(b).

This looks like a logarithm -- `a^m*a^n=a^(m+n)` . Thus the function `f(x)=int_1^x 1/x dx` was named the natural logarithm. Since it is continuous and increasing there is a c such that f(c)=1; that c is of course `e~~2.718281828...`

You can use substitution:

`int-2/(2x+1)dx=|(t=2x+1),(dt=2dx)|=int-dt/t=-ln|t|=-ln|2x+1|`

So your solution isn't `-2ln|2x+1|` but `-ln|2x+1|`.

However ther is an easier way to solve this type of integral. You can use the following formula:

`int (f'(x))/(f(x))dx=ln|f(x)|` **(1)**

Let's prove the formula:

`(ln|f(x)|)'=(f'(x))/(f(x))` **(2)**

Now to get **(1)** we only integrate both sides of **(2)**` ` and our proof is done.

So in your case `(2x+1)'=2` hence by using formula (1) we get `int-2/(2x+1)dx=-ln|2x+1|`.