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Problem.1) Evaluate $$ \int \frac{x^5}{\sqrt{x^2 + 1}} \, \mathrm{d}x $$

I'm still learning how to integrate, watching the GOAT of calculus, BlackPenRedPen. it would be quiet helpful if anyone could help me understand this. Thanks.

Sangchul Lee
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Ian Hendrick Pugoy
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    Welcome to [math.se] SE. Take a [tour]. You'll find that simple "Here's the statement of my question, solve it for me" posts will be poorly received. What is better is for you to add context (with an [edit]): What you understand about the problem, what you've tried so far, etc.; something both to show you are part of the learning experience and to help us guide you to the appropriate help. You can consult this link for further guidance. – Another User Dec 09 '23 at 15:03
  • Hint: $u=x^2+1$. – Deif Dec 09 '23 at 15:15
  • https://math.stackexchange.com/questions/1662305/trig-substitution and https://math.stackexchange.com/questions/997462/evaluating-int-fracx3-left-sqrt4x29-right3-dx-using-a-trigonomet – lab bhattacharjee Dec 09 '23 at 16:42
  • HINT $5 =2 *2 +1$ so $x^5= (x^2)^2 x$ sub $u =x^2$, $du= 2x dx$ $x^5 dx = 0.5 u^2 du$ – pie Dec 09 '23 at 18:08

2 Answers2

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I would use a trig substitution. Set up a triangle with a hypotenuse of length $\sqrt{x^2+1}$, adjacent leg of length $1$, and opposite leg of length $x$. Then, you have a couple useful substitutions available: $$\sec\theta = \sqrt{x^2+1}\\ \tan\theta = x \implies \sec^2\theta d\theta = dx$$ Now, simply substitute as follows: $$\int\frac{x^5}{\sqrt{x^2+1}}dx = \int\frac{\tan^5\theta}{sec\theta}\cdot\sec^2\theta d\theta$$ Then, simplify by cancelling $\sec\theta$ and now you can solve the integral with fairly basic trig techniques:

$$=\int tan^4\theta\cdot \tan\theta\sec\theta d\theta\ =\int (\sec^2\theta-1)^2\cdot \tan\theta\sec\theta d\theta\ =\int(\sec^4\theta -2\sec^2\theta +1)\cdot\tan\theta\sec\theta d\theta$$ Then $u=\sec\theta \implies du = \sec\theta\tan\theta d\theta$ so $$= \int (u^4-2u^2+1)du\ = \frac{1}{5}u^5-\frac{2}{3}u^3+u+C\$$ And $u=\sec\theta = \sqrt{x^2+1}$ so the final answer is $$\frac{1}{5}(x^2+1)^{5/2}-\frac{2}{3}(x^2+1)^{3/2}+(x^2+1)^{1/2}+C$$

logan-edwards
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Attempt the question for yourself, and if you have done so then here is an answer with working:

$$\int \frac{x^5}{\sqrt{x^2+1}}\,dx$$ Let's use the substitution $u = x^2+1$. Then we have $dx = \frac{1}{2x}\,du$, and the integral becomes $$\frac{1}{2}\int \frac{x^4}{\sqrt{u}}\,du$$ $u = x^2 +1 \quad \Rightarrow x^4 = (u-1)^2$ $$\frac{1}{2}\int \frac{(u-1)^2}{\sqrt{u}}\,du$$ $$= \frac{1}{2}\int \frac{u^2-2u+1}{\sqrt{u}}\,du$$ $$= \frac{1}{2}\int u^{2-1/2}-2u^{1-1/2}+u^{-1/2}\,du$$ $$= \frac{1}{2}\int u^{3/4}-2u^{1/2}+u^{-1/2}\,du$$ $$= \frac{1}{2}\left(\frac{2}{5}u^{5/2}-\frac{4}{3}u^{3/2}+2u^{1/2}\right)+C$$ $$= \frac{1}{5}u^{5/2}-\frac{2}{3}u^{3/2}+u^{1/2}+C$$ which after plugging $u = x^2 +1$ back in, we get $$\int \frac{x^5}{\sqrt{x^2+1}}\,dx = \frac{(x^2 +1)^{5/2}}{5}-\frac{2}{3}(x^2+1)^{3/2}+(x^2+1)^{1/2}+C$$

Mako
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    Slight remark: you wrote that $\mathrm{d}x=\frac{1}{2}\mathrm{d}u$, but it should be $x,\mathrm{d}x=\frac{1}{2}\mathrm{d}u$ (which you did you) – Lorago Dec 09 '23 at 17:35
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    Thats a typo. Thanks for pointing it out – Mako Dec 09 '23 at 17:36