I am trying to find the integral to $$\frac{3x}{(x+1)^4}$$
My question is, I have been attempting to solve the answer using integration by parts my workings
I can't arrive at the given answer, and online calculators seems to be pushing for me to substitute $$u = x + 1$$
It seems that I can't naively do the integration by parts for this… is there a reason why?
Best Answer
Of course you can use integration by parts! Your calculations are correct, except that $1/2$ mysteriously became $3/2$ in the last step, and that you forgot to add $C$.
Also, when you multiply by something with a minus sign, you need to write it as $x (-\tfrac13)(x+1)^{-3}$, not $x-\tfrac13(x+1)^{-3}$ which means something completely different.
To compare your answer with the one from symbolab, note that $$ \frac{x}{(x+1)^3} = \frac{x+1-1}{(x+1)^3} = \frac{1}{(x+1)^2} - \frac{1}{(x+1)^3} . $$