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I'm currently studying for my calc exam, and i've stumbled across a rather odd problem(at least for me). I've been doing integrals non stop for 2 weeks now and I haven't seen anything like this, so I would like to know if my approach is correct. I feel like i'm not doing it correctly since my answer conflicts with my professor's answer key. It is as follows:

$$\int\sqrt{x^4+x^7}\;dx\;\; or \int(x^4+x^7)^\frac{1}{2}\;dx$$

Since i've been doing(mostly) complex trig sub and integration by parts, I didn't immediately know what to do. I decided to convert the integral to $\int(x^4+x^7)^1/2$ and multiply the exponents:

$$\int\sqrt{x^4+x^7} = \int(x^2+x^\frac{7}{2})\;dx$$

Then I use basic integration to yield:

$$\frac13x^3+\frac29x^\frac{9}{2}+\;C$$

Taking the derivative to check:

$$\frac {d}{dx}(\frac 13x^3 + \frac29x^\frac{9}{2}) = x^2+x^\frac{7}{2}$$

Seems to give me what I started with, but my answer key has this as the answer: $$\frac29(1+x^3)^\frac32+C$$

I can see some similarities to my answer, but it makes me feel like I made a mistake in my technique. Symbolab isn't capable of computing this integral for some reason, and WolframAlpha gives an answer far, far different then either of the integrals I(or my professor) has. Any input would be greatly appreciated as I just want to be as prepare for my exam as much as possible. Thanks!

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    $\begingroup$ $\sqrt{a+b}\neq \sqrt{a}+\sqrt{b}.$ Try to factor $x^4+x^7$ as $x^4(1+x^3).$ $\endgroup$ Jun 15 '16 at 9:33
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    $\begingroup$ Thank you for pointing that out, I felt like I was making a critical algebra mistake. I've been at it for so long, I may need to call it quits for the night. $\endgroup$ Jun 15 '16 at 9:41
  • $\begingroup$ @FuegoJohnson You did what i always do at late night. $\endgroup$
    – user312097
    Jun 15 '16 at 9:50
  • $\begingroup$ @ritwiksinha I know, right? I get tired and silly mistakes start to happen. I never like closing on a problem I didn't get first try though, I'm stubborn :p $\endgroup$ Jun 15 '16 at 9:53
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    $\begingroup$ @FuegoJohnson I can't sleep, if i start a problem and couldn't finish it. Then i start changing the laws of mathematics just to finish it. $\endgroup$
    – user312097
    Jun 15 '16 at 9:56
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HINT

Take out $x^4$ common from squareroot and then put $x^3=t$

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  • $\begingroup$ Thank you, I reached the correct answer. Not only did I make a critical algebra mistake, but I think I also complicated things way more than I needed to. I'm very tired, but I will not make this mistake again for my exam. Thanks again for your input. $\endgroup$ Jun 15 '16 at 9:49
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    $\begingroup$ All the best for exam $\endgroup$
    – Gathdi
    Jun 15 '16 at 9:50
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We have: $\displaystyle\int{\sqrt{x^{4}+x^{7}}}dx$

Let's begin by taking an $x^{4}$ in common:

$=\displaystyle\int{\sqrt{x^{4}(1+x^{3})}}dx$

$=\displaystyle\int{\sqrt{x^{4}}\sqrt{(1+x^{3})}}dx$

$=\displaystyle\int{x^{2}\sqrt{(1+x^{3})}}dx$

Then, let's try using integration by substitution.

Let $u=1+x^{3} \Rightarrow du=3x^{2}dx$:

$=\dfrac{1}{3}\displaystyle\int{3x^{2}\sqrt{u}}dx$

$=\dfrac{1}{3}\displaystyle\int{\sqrt{u}}du$

$=\dfrac{1}{3}\displaystyle\int{u^{\frac{1}{2}}}du$

$=\dfrac{1}{3}\bigg(\dfrac{u^{\frac{1}{2}+1}}{\frac{1}{2}+1}\bigg)+C$

$=\dfrac{1}{3}\bigg(\dfrac{u^{\frac{3}{2}}}{\frac{3}{2}}\bigg)$

$=\dfrac{2}{9}\sqrt[3]{u^{2}}+C$

Now, replacing $u$ with $1+x^{3}$:

$=\dfrac{2}{9}\sqrt[3]{\big(1+x^{3}\big)^{2}}+C$

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  • $\begingroup$ Thank you for the in depth answer. I was able to solve it based off the hint @Gathdi made but this is a nice breakdown for others who may be in a similar situation I was. $\endgroup$ Jun 15 '16 at 19:17
  • $\begingroup$ No worries, hope you do well on the calc test $\endgroup$ Jun 19 '16 at 18:55
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    $\begingroup$ Thanks! I actually think i did quite well, all the help from you folks prepared me really well. I'll know tomorrow $\endgroup$ Jun 20 '16 at 19:42

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