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I'm trying to solve this (not homework, if it matters), and both u-substitution and integration by parts are both yielding two different answers. Where am I going wrong?

Equation: $$\int \frac{(4x^3)}{(x^4+7)}dx$$

u-substitution answer: $$=\ln\big|(x^4+7)\big|+C$$

integration by parts answer: $$=\int4x^3*(x^4+7)^{-1}dx$$ $$=4x^3*\ln\big|x^4+7\big|-\int 12x^2*(x^4+7)^{-1}dx$$ $$=4x^3*\ln\big|x^4+7\big|-(12x^2*ln\big|x^4+7\big|-\int 24x*(x^4+7)^{-1}dx)$$ $$=4x^3*\ln\big|x^4+7\big|-(12x^2*ln\big|x^4+7\big|-24x*ln\big|x^4+7\big|-\int 24(x^4+7)^{-1}dx)$$

$$= 4x^3*\ln\big|x^4+7\big|-(12x^2*\ln\big|x^4+7\big|-(24x\ln\big|x^4+7\big|-24\ln\big|x^4+7\big|))$$ $$=(4x^3-12x^2+24x-24)(\ln\big|x^4+7\big|)$$

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  • $\begingroup$ The two results are false. $\endgroup$ – JJacquelin Feb 14 '14 at 16:33
  • $\begingroup$ Both are wrong. You're making a mistake on passing from rational functions to a log primitive. The integral should be just $log(x^4+7)$, since differentiating it comes out as $f(x) '/f(x) $ $\endgroup$ – Felipe Jacob Feb 14 '14 at 16:33
  • $\begingroup$ I just updated the u-substitution answer, still not sure on the int-by-parts, I'll update with my work $\endgroup$ – Josh F Feb 14 '14 at 16:37
  • $\begingroup$ [NB: Try using $$\begin{align}A&=B \\ &=C \\ &\dots\end{align}$$ for $$\begin{align}A&=B \\ &=C \\ &=\dots\end{align}$$] $\endgroup$ – Shaun Feb 14 '14 at 16:50
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Just to answer your question about why using integration by parts allegedly leads to a different solution: In your work on integration by parts, you made a mistake at the start; you seem to have let $ u =4x^3$, $\,dv = (x^4 + 7)^{-1}\,dx$. The problem is in your conclusion that $uv = 4x^3 \ln(x^4+ 7)$.

It is NOT true that in integrating $dv$, we obtain $v = \ln(x^4 + 7)$. In order to integrate $dv$, you're attempting to integrate $$\int\dfrac{dx}{(x^4 +7)}\neq \ln(x^4 + 7) +c$$ We need $u = 4x^3$ to be back in the integrand to obtain $$\int \dfrac{(x^4 + 7)'}{x^4 + 7}\,dx = \ln(x^4 + 7) + c$$

In short, you're back to the original integral.

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I don't understand it. For $u = x^4+7, du = 4x^3dx$ so $$ \int \frac{4x^3}{x^4+7} dx = \int du/u = \ln |x^4+7| + C. $$

Show work for your by parts results and it will become clear where the error is...

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  • $\begingroup$ Quickest answer with no need for repetition by others! + (In all fairness to other answerers, they were also quick, and probably in the process of answering while you posted first.) $\endgroup$ – Namaste Feb 14 '14 at 16:33
  • $\begingroup$ thanks! I mis-typed the u-substitution answer and re-wrote it as what you have. $\endgroup$ – Josh F Feb 14 '14 at 16:47
  • $\begingroup$ But with the integration by parts answer, I wrote out my work, but I can't quite see why it doesn't work. Thoughts? $\endgroup$ – Josh F Feb 14 '14 at 16:47
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I know you are asking about a specific math problem, however, I ran into the same issue with finding the integral of $$ \int x(x-2)^5 dx.$$ The two answers I received were $$\frac{(x-2)^7}{7}+\frac{x-2^6}{3}+C$$ by using the Chain Rule and $$\frac{x (x-2)^6}{6}-\frac{(x-2)^7}{42} +C$$ using the Product Rule (integration by parts or tabular method). At first I thought they were different answers but I expanded both answers in my TI-89 (could be done by hand but will take a minute). Then I factored each expanded answer and received $$\frac{(x-2)^6 (3x+1)}{21}+C$$ for both answers. So through that route I figured out both yield the same answer but in different algebraic form.

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  • $\begingroup$ I don't think that this will help the OP to solve his problem. $\endgroup$ – miracle173 May 10 '17 at 3:42
  • $\begingroup$ It could if OP hadn't made a mistake, then spent time changing the final forms to resemble each other. $\endgroup$ – JDG Feb 13 at 3:11
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The correct result is: $$\int\frac{4x^3}{x^4+7}dx=\ln|x^4+7|+c$$.

With substitution: $u=x^4+7$ from which $du=4x^3dx$ hence the integral become: $$\int\frac{du}{u}=\ln|u|+c$$.

Integration by parts it's not raccomanded in this case.

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Both answers are wrong, I believe. Letting $u = x^4 + 7$, you get $du = 4x^3 ~dx$, sop the first integral becomes $$ \int \frac{1}{u} du = \ln u = \ln (x^4 + 7). $$

For the second, it's not at all clear to me which parts of the integral you've chosen as $u$ and $dv$ for your integration by parts.

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