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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 $$u = x + 1$$

https://www.symbolab.com/solver/by-parts-integration-calculator/by%20parts%20%5Cint%20%5Cfrac%7B3x%7D%7B%5Cleft(1%2Bx%5Cright)%5E%7B4%7D%7Ddx?or=input

It seems that I can't naively do the integration by parts for this... is there a reason why?

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  • $\begingroup$ See the answer of the following: math.stackexchange.com/questions/538654/… $\endgroup$
    – nmasanta
    Commented Mar 21, 2022 at 11:08
  • $\begingroup$ hi, thanks for this. This is helpful for tips but it doesn't explain why integration by parts cannot lead me to the same answer. $\endgroup$ Commented Mar 21, 2022 at 11:12
  • $\begingroup$ This is quite informal, but generally you can understand integration by parts as the inverse to the product rule while substitution is the inverse of the chain rule. You can hence not expect that integration by parts works whenever you want – it can only work if differentiating the (unknown) integral function involves the product rule. $\endgroup$
    – junjios
    Commented Mar 21, 2022 at 11:12
  • $\begingroup$ @msgcas is this why then people always recommend substitution? wish this was actually mentioned somewhere in my materials..... I had thought they work bothways! $\endgroup$ Commented Mar 21, 2022 at 11:18
  • $\begingroup$ Even without the substitution, writing the numerator as $3x=3x+3-3$ will help and separate the single fraction into two easily integrable fractions. In fact this is a very basic application of the method of partial fractions, which means that when you have only linear factors in the denominator you can rewrite your integrand as a sum so that you have constants in every numerator and simple powers in every denominator. Integration by parts is less simple and more prone to error. $\endgroup$ Commented Mar 21, 2022 at 11:18

2 Answers 2

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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 $$ \frac{x}{(x+1)^3} = \frac{x+1-1}{(x+1)^3} = \frac{1}{(x+1)^2} - \frac{1}{(x+1)^3} . $$

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  • $\begingroup$ Hans, appreciate you taking the time but I still can't see how to get to Symbolabs answer from my final step. It seems that you agree with my answer (other than the 3/2 and the lack of C). For our purposes , if we correct the mistake and add C then $$\frac{-x}{(x+1)^3} - \frac{1}{2(x+1)^2} + C $$ it is still vastly different from Symbolabs and your answer comparision! $\endgroup$ Commented Mar 21, 2022 at 11:41
  • $\begingroup$ So what happens if you rewrite the first term in the way that I suggested...? $\endgroup$ Commented Mar 21, 2022 at 11:48
  • $\begingroup$ (Or, alternatively, just take your answer minus theirs and simplify the whole shebang; you'll get zero.) $\endgroup$ Commented Mar 21, 2022 at 11:51
  • $\begingroup$ Hans, is it possible to show me what exactly u meant by the last 2 lines? I am genuinely unable to follow the thought process and am trying my hardest to solve using your suggestion. my trouble being $$-x$$ rather than the $$x$$ that you have there. $\endgroup$ Commented Mar 21, 2022 at 12:03
  • $\begingroup$ $$\dfrac{-x}{(x+1)^3} = - \dfrac{x}{(x+1)^3} = - \biggl( \quad \cdots \quad \biggr)$$ $\endgroup$ Commented Mar 21, 2022 at 12:26
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You can use integration by parts without substitution as below: $$ \begin{aligned} \int \frac{3 x}{(x+1)^{4}} d x &=-\int x d\left(\frac{1}{(x+1)^{3}}\right) \\ &=-\frac{x}{(x+1)^{3}}+\int \frac{d x}{(x+1)^{3}} \\ &=-\frac{x}{(x+1)^{3}}-\frac{1}{2(x+1)^{2}}+C \\ &=-\frac{3 x+1}{2(x+1)^{3}}+C \end{aligned} $$

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  • $\begingroup$ thank you for this. $\endgroup$ Commented Mar 22, 2022 at 11:03

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