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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?

dragostiadintei
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  • See the answer of the following: https://math.stackexchange.com/questions/538654/when-to-do-u-substitution-and-when-to-integrate-by-parts – nmasanta Mar 21 '22 at 11:08
  • 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. – dragostiadintei Mar 21 '22 at 11:12
  • 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. – junjios Mar 21 '22 at 11:12
  • @msgcas is this why then people always recommend substitution?

    wish this was actually mentioned somewhere in my materials..... I had thought they work bothways!

     – dragostiadintei Mar 21 '22 at 11:18
  • 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. – Mark Bennet Mar 21 '22 at 11:18
  • Note also that it is a common hazard, when comparing two methods of integration, that the answers you get with the different methods differ by a constant - and sometimes that constant is hidden within a formula, rather than visible on the surface. – Mark Bennet Mar 21 '22 at 11:20
  • @MarkBennet I am aware of the difference by the constant, u are right in that; I just had attempted to use the site to show me the steps I may have neglected in this question.

    It seems that integration by parts is not really useful to tackle all problems; which is something new I learnt today.

     – dragostiadintei Mar 21 '22 at 11:23
  • thank u all for your time. – dragostiadintei Mar 21 '22 at 11:26
  • I definitely wouldn't always recommend substitution. I was just saying that it really depends on your problem which one works and which one doesn't. Consider e.g. $f(x):=x(\ln(x)-1)$. Via the product rule you find $f'(x)=\ln(x)$. Conversely, given $f'$ you may find $f$ via integration by parts. On the other hand, if $g(x):=e^{x^2}$, then you find with the chain rule: $g'(x)=2xe^{x^2}$. Given $g'$ you may find $g$ using the substitution $y:=x^2$. – junjios Mar 21 '22 at 11:27
  • @msgcas this is helpful, and thanks for the suggestion.

    i think this means I need to get better at pattern recognition and getting faster at this over arbitrarily deciding to use which methods, it could be that oen is more intuitive.

     – dragostiadintei Mar 21 '22 at 11:33
  • It can indeed be quite difficult to see which method will help you. Usually, I think first about integration by parts because imo it's a lot easier to see whether it'll get you anywhere than with substitution. In practice, if we want to integrate a product of two functions, say $u\cdot v$ you would want that at least one of $u$ and $v$ can be integrated easily, say you see e.g. immediately that $u=w'$. Then, often integration by parts will help you if $wv'$ "looks a lot easier" than $uv$. In my example above we'd have $u(x)=1$, $v(x)=\ln(x)$ and $w(x)=x$. Then, $w(x)v'(x)=1$ looks very simple. – junjios Mar 21 '22 at 11:48
  • @msgcas thanks for the tip! – dragostiadintei Mar 21 '22 at 11:49

2 Answers2

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

Hans Lundmark
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  • 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!

     – dragostiadintei Mar 21 '22 at 11:41
  • So what happens if you rewrite the first term in the way that I suggested...? – Hans Lundmark Mar 21 '22 at 11:48
  • (Or, alternatively, just take your answer minus theirs and simplify the whole shebang; you'll get zero.) – Hans Lundmark Mar 21 '22 at 11:51
  • 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.

     – dragostiadintei Mar 21 '22 at 12:03
  • $$\dfrac{-x}{(x+1)^3} = - \dfrac{x}{(x+1)^3} = - \biggl( \quad \cdots \quad \biggr)$$ – Hans Lundmark Mar 21 '22 at 12:26
  • https://imgur.com/a/Do7SieJ

    i think this is it? I solved the answer in that case.

    not going to lie i lost a few hairs doing this but Hans thanks for showing me the way... (BTW: realised i had the denominator jumbled halfway in the question should be $(1+x)$but the core confusion still stands)

     – dragostiadintei Mar 21 '22 at 12:29
  • Yes, exactly. Now you're subtracting $1+\tfrac12 = \tfrac32$ times $1/(1+x)^2$. – Hans Lundmark Mar 21 '22 at 15:14
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You can use integration by parts without substitution as below: $$ \begin{aligned} \int \frac{3 x}{(x+1)^{4}} d x &=-\int x d\left(\frac{1}{(x+1)^{3}}\right) \\ &=-\frac{x}{(x+1)^{3}}+\int \frac{d x}{(x+1)^{3}} \\ &=-\frac{x}{(x+1)^{3}}-\frac{1}{2(x+1)^{2}}+C \\ &=-\frac{3 x+1}{2(x+1)^{3}}+C \end{aligned} $$

Lai
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