0
$\begingroup$

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.

$\endgroup$
2
$\begingroup$

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 $2+x^7$.

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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.