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I have the following 2 questions to integrate as part of my practice.

(i) $displaystyle intfrac1{x(x+2)}dx$

(ii) $displaystyle intfrac1{e^x+2}dx$

and I have tried it myself but I am not getting the answer required

(i) $displaystyle frac12ln x-frac12ln(x+2)+C$

(ii) $displaystyle frac12x-frac12ln(e^x+2)+C$

Below are my workings

(i) $displaystyle intfrac1{x(x+2)}dx;=;intfrac1{x^2+2x}dx;=;frac{ln(x^2+2x)}{2x+2} +C$

(ii) $displaystyle intfrac1{e^x+2}dx;=frac{ln(e^x+2)}{e^x} +C$

May I know what am I doing wrong such that my answer differs?

For i):

With the constant rule 'trick', what you are effectively doing is:
$$int frac{1}{x^2+2x} dx$$

When $u = x^2+2x, du = 2x+2 dx, dx = frac{du}{2x+2}$, we have:

$$int frac{1}{u(2x+2)} du, $$

and since you cannot get the $2x+2$ part to be in terms of $u$, this 'trick' does not work.

What you can do is to use partial fractions:
$$frac{A}{x} + frac{B}{x+2} = frac{1}{x(x+2)}$$
$$A(x+2) + Bx = 1 tag{1}label{eq1}$$
$$(A+B)x + 2A = 1 tag{2}label{eq2}$$

This is an identity – it will hold for any $x$. So when $x = 0, A = frac{1}{2}$ from equation $2$, and when $x = -2$, $B = -frac{1}{2}$ from equation $1$.

Can you integrate it now?

For i) use partial fractions; note that $$frac{1}{x(x+2)}=frac{frac 12}{x} -frac{frac12}{x+2}$$ So your integral becomes:
$$int{frac{1}{x(x+2)}dx}=frac12bigg[int frac 1x dx-int frac{1}{x+2}dxbigg]$$

For the second, substitute $$u=e^xto dx = frac{du}{u}$$

Then your integral becomes:

$$intfrac{1}{e^x+2}dx=int{frac{1}{u(u+2)}du}$$ which we just solved in i)

The substitution rule is $$int f(g(x)) , g'(x) ,{rm d}x = int f(u) ,{rm d}u.$$ In particular, when $f(x) = 1/x$, $$int frac{g'(x)}{g(x)},{rm d}x = int frac{{rm d}u}{u} = log u + C = log(g(x)) + C.$$

What you are trying to do is to "dividing both sides by the derivative of $g$": $$int frac{1}{g(x)},{rm d}x = frac{log(g(x))}{g'(x)} + C. tag{OP's wrong method}$$ As the first comment suggests, differentiating both sides with respect to $x$ allows you to see the error.