Question

How do you evaluate the definite integral `int x/(x+1)` from `[0,1]`?

Answer 1

I found: `1-ln(2)`

Explanation:

We can solve first the indefinite integral by manipulating a bit the integrand:
`int(x/(x+1))dx=-int(1/(x+1)-1)dx=-[int1/(x+1)dx-intdx]=-[ln(x+1)-x]`
evaluated between `x=0` and `x=1`:

`=-[ln(2)-1-ln(1)+0]=1-ln(2)`

Answer 2

We have that

`int (x+1-1)/(x+1)*dx=int (1-1/(x+1))dx=x-ln(x+1)+c`

Hence

`int_0^1 x/(x+1)dx=[x-ln(x+1)]_0^1=1-ln2=0.30685`

Answer 3

The answer is `=(1-ln2)`

Explanation:

We do this by substitution, `u=x+1`

Then `du=dx`

So, `int(xdx)/(x+1)=int((u-1)du)/u`

`=int(1-1/u)du=u-lnu +C`

`=(x+1)-ln(x+1)+C`

Therefore,

`int_0^1(xdx)/(x+1)=[(x+1)-ln(x+1)] _0^1`

`=2-ln2-1-0=(1-ln2)`