The answer I have which I'm sure is wrong is $(3x-2)\ln(\left \vert x+1 \right \vert + \text{constant}$
I let $x + 1$ be $u$ and $du$ would be $1$. But I didn't know how to get $1$ on top so I just pulled the numerator $(3x-2)$ to the outside and left a $1$ on top.
Can I do that?
$\endgroup$1 Answer
$\begingroup$Yes, you are wrong. We have $$\dfrac{3x-2}{x+1} = \dfrac{3(x+1)-5}{x+1} = 3 - \dfrac5{x+1}$$ Now integrating this gives us $$3x-5\ln(\left\vert x+1 \right \vert) + \text{constant}$$
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