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Consider the following question:

Find the antiderivative of $f(x) = 6x \, (x^2+1)^5$.

I have been using 2 notations and I would like to combine them.

  1. Notice that $[x^2 + 1]' = 2x$, so $6x \, dx = 3 d(x^2+1)$

  2. Take $(x^2 + 1) = u$, then $6x$ is $3\,u'$.

I would like to combine these, but I'm wondering if it would still be correct:

Take $[x^2 +1] = u$, then $6x = 3 \, du$. This would save me a lot of time on tests, and also would make it easier, however with calculus I am never sure if my notation is correct.

Is this correct, and if not; what would the optimal way to combine these methods, to be as concise as possible, be?

By the way, this would be how I would do the problem:

Take $[x^2 +1] = u$, then $6x = 3 \, du$.

$6x(x^2+1)^5 = 3du \cdot u^5 = 3u^5 du = d \dfrac{1}{2}u^6 = d \dfrac{1}{2}(x^2+1)^6$.

$F(x) = \dfrac{1}{2} (x^2+1)^6 + c$.

ByeByeYa
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  • Don't forget the $dx$ on the left hand side. Otherwise, it looks good. – Adriano Jun 26 '13 at 17:20
  • @Adriano You mean after $6x$? I have noticed that in my textbook too, why is that neccesary? – ByeByeYa Jun 26 '13 at 17:47
  • It's because we will be integrating it. One way to think about it is to abuse the notation a bit and treat $dy/dx$ as a separable fraction (mathematicians hate doing this, but it's a good mnemonic technique). By "cross multiplying", this leads to: $$ u=x^2+1 \implies \dfrac{du}{dx} = 2x \implies du=2xdx \implies 3du = 6xdx $$ – Adriano Jun 26 '13 at 17:51
  • @Adriano Thanks, that helps! – ByeByeYa Jun 26 '13 at 18:11

1 Answers1

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Yes, that will work: $$u = x^2 + 1 \;\implies\; du = 2x\,dx,\;\text{so}\; 6x\,dx = 3 du$$

and the resulting integral will be $$\int 6x(x^2 + 1)^5 \,dx = \int u^5 \,(3 du) = 3\int u^5\,du$$

Now we just integrate with respect to $u$:

$$3\int u^5\,du = 3\left[\frac{u^6}{6}\right] + C$$

Back substituting gives us $$\frac 12(u^6) + C = \frac 12(x^2 + 1)^6 + C$$

amWhy
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  • Can you please check if my answer to the question is correct too? The main part which bothers me is the first line, because I don't know if you just have to put $6x(x^2+1)^5 = ..$ or $\displaystyle \int6x(x^2+1)^5 dx = ..$ – ByeByeYa Jun 26 '13 at 17:20
  • You should use integral notation to denote taking the antiderivative of $6x(x^2 + 1)^5 \implies$ evaluating $$\int 6x(x^2 + 1)^5 ,dx$$ – amWhy Jun 26 '13 at 17:23
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    And can we add another step: $.. = 3 \displaystyle \int u^5 , du = \dfrac{1}{2}u^6$? – ByeByeYa Jun 26 '13 at 17:26
  • Yes, exactly, then you can back-substitute, once you've integrated, replacing each occurrence of $u$ (here, one occurrence) with $x^2 + 1$. – amWhy Jun 26 '13 at 17:28
  • Wow, sorry for the plethora of questions, but the very last one; If, as I did in my previous comment, you integrate and hence the integral sign is removed, does that mean the $+c$ has to be added? Does the absence of the integral sign mean that you know must add the $+c$ for it to be correct? – ByeByeYa Jun 26 '13 at 17:30
  • Yes, the absence of the integral sign indicates that we have integrated the integrand, resulting in F(x) + C, and with indefinite integrals, we need to recognize the solution includes a family of all such functions $F(x) + C$ which vary by only a constant. (If you turn back and differentiate $F(x) + C$ to obtain the original "prior to integrating" function $f(x)$, the constant term $C$ vanishes, since $\frac d{dx}(\text{constant}) = 0$.) – amWhy Jun 26 '13 at 17:33
  • @amWhy: More nice feedback! +1 – Amzoti Jun 27 '13 at 00:39