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For this problem I did the integration by parts and got that answer. But for some reason it is still wrong. Does anyone get any different answers? Any hints or suggestions would be of great help! Thanks!

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You need to "cancel" an $x$ in that 2nd fraction –  imranfat Feb 24 at 17:12
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4 Answers

up vote 2 down vote accepted

You would have to make $u= ln(x)$ and $du= (1/x)$. Then you would make $dv= x^3$ and v would then = $x^4/4$. Using the integration by parts formula, which is the integration of u dv = uv- integration of $vdu$. Substitute in your variables and your answer should be $(ln(x) x^4/4) - (x^4/16) + C$.

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Yah I saw the silly mistake that I made. Thank you for being clear and explaining it all. –  AlecLeonK Feb 24 at 17:19
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By $u=\ln x$ then $u'=\frac 1x$ and $v'=x^3$ then $v=\frac14 x^4$ so

$$\int x^3\ln xdx=uv-\int u' vdx=\frac 14x^4\ln x-\frac14\int x^3 dx=\frac 14x^4\ln x-\frac1{16}x^4+C$$

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Got it...silly me! –  amWhy Feb 24 at 17:17
    
you're welcome amWhy:-) –  Sami Ben Romdhane Feb 24 at 17:18
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$$\int x^3\ln x\,\text{d}x=\int\ln x\cdot\text{d}\left(\dfrac{x^4}{4}\right)=\dfrac{x^4}{4}\ln x-\int \dfrac{x^4}{4}\cdot\text{d}(\ln x)=\dfrac{x^4}{4}\ln x-\int\, \dfrac{x^3}{4}\text{d}x=\cdots$$

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you integrated $\frac{1}{4}\frac{x^4}{x}$ and got $\frac{1}{20} \frac{x^5}{x}$ which is not right. Cancel the $x$ from the denominator of the first expression, then integrate.

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What a silly mistake, thank you for recognizing it! I got it right. –  AlecLeonK Feb 24 at 17:15
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