In my book they make the point that: "The method of substitution cannot be force to work".

They then go on to say that: "there is no substitution that will do much good with the integral $\int x(2+x^7)^{\frac{1}{5}}dx $ "

However, the integral $\int x^6(2+x^7)^{\frac{1}{5}}dx $ does apparently yield a substitution for $2+x^7$. Where $u=2+x^7$

Why is this? What distinguishes the two, obviously the order of the exponent $x$ and $x^6$ does something. But I am unsure about what.


Consider $$z=2+x^7$$ $$dz=7x^6dx\\\text{or}$$ $$x^6dx=dz/7$$ This means, $$\int x^6(2+x^7)^{1/5}dx=\int\frac{1}{7} z^{1/5}dz\\\text{which can easily be calculated to be equal to }\frac{5z^{6/5}}{42}.$$ Note that nothing of this sort could happen in the case $x^1$, since it won't come out to be of the form of the differentiation of $2+x^7$.

  • 1
    $\begingroup$ I see. Thank you! $\endgroup$
    – oxodo
    Dec 24 '18 at 0:16

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