0
$\begingroup$

I'm trying to integrate

$$\int(5x^4+1)\ln(x^5+x)dx$$

When I use an online calculator (it first uses substitution and then integration by parts) I get the answer

$$(x^5+x)\ln(x^5+x)-x^5-x$$

but when I try a different approach (without substitution only with integration by parts) I get

$$(x^5+x)\ln(x^5+x)-x$$

(I change the variable of $dx$ to $x^5+x$)

$\endgroup$
  • 1
    $\begingroup$ The first is correct modulo a constant of integration, the second is wrong. Without any more info, this is because you didn't do the second method correctly. $\endgroup$ – Paul Jan 4 '18 at 14:54
  • 1
    $\begingroup$ Use Latex to type your question. Put all the math parts inside two dollar signs $$ and use the command \int_{}^{} to display your intergral $\endgroup$ – pureundergrad Jan 4 '18 at 14:54
  • $\begingroup$ Is that integral supposed to be $\int(5x^4+1)\ln(x^5+x)dx$ ? $\endgroup$ – PM 2Ring Jan 4 '18 at 15:00
  • $\begingroup$ @PM2Ring yes that's it $\endgroup$ – john Jan 4 '18 at 15:02
1
$\begingroup$

It seem's that you forgot something since diferentiation of $(x^5+x)\ln(x^5+x)-x$ gives: \begin{align*} D_x\left[(x^5+x)\ln(x^5+x)-x\right]&=(5x^4+1)\ln(x^5+x)+5x^4+1-1\\ &=(5x^4+1)\ln(x^5+x)+5x^4 \end{align*} And in the expression $$(x^5+x)\ln(x^5+x)-x$$ the last $x$ must be $x^5+x$ because the change that you used in order to integrate, i.e. $x^5+x$ instead of $x$.

$\endgroup$
  • $\begingroup$ oh it seems I didn't calculate the first derivate the right way... didn't see it was a function of a function... $\endgroup$ – john Jan 4 '18 at 15:19
0
$\begingroup$

Since your integral looks like $\int g’\ln(g)$ you just need to find a primitive of $\ln( x)$, namely $x\cdot\ln(x)-x$, and just plug in $x^5+x$ for $g$.

$\endgroup$

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.