Can you solve this week's POTW: Integrating a Complex Expression? May 25th, 2020

[anemone]

Here is this week's POTW:

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Find $$\int \left(x^{10}+\sqrt{1+x^{20}}\right)^{^{\Large\frac{21}{10}}}\,dx$$ where $x\in \mathbb{R}$.

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[anemone]

No one answered last week's POTW. (Sadface) However, you can find the suggested solution of other below:

Spoiler: Solution of other

$$\begin{align*}\int \left(x^{10}+\sqrt{1+x^{20}}\right)^{^{\Large\frac{21}{10}}}\,dx&=\int \left(\dfrac{(1+x^{20})-x^{20}}{\sqrt{1+x^{20}}-x^{10}}\right)^{^{\Large\frac{21}{10}}}\,dx\\&=\int (\sqrt{1+x^{20}}-x^{10})^{^{\Large-\frac{21}{10}}}\,dx\\&=\int (\sqrt{1+x^{-20}}-1)^{^{\Large-\frac{21}{10}}}\cdot x^{-21}\,dx\end{align*}$$

Let $u=\sqrt{1+x^{-20}}-1$. We then have

$x^{-20}=u^2+2u\\-20x^{-21}dx=(2u+2)du$

$$\begin{align*} \therefore \int \left(x^{10}+\sqrt{1+x^{20}}\right)^{^{\Large\frac{21}{10}}}\,dx&=\int u^{\Large-\frac{21}{10}}\left(-\dfrac{1}{10}(u+1)\right)\,du\\&=-\dfrac{1}{10} \int (u^{\Large-\frac{1}{20}}+u^{\Large-\frac{21}{20}})\,du\\&=-\dfrac{1}{10}\left(\dfrac{20}{19}u^{\Large\frac{19}{20}}-20u^{\Large-\frac{1}{20}}\right)+C\\&=-\dfrac{2}{19}(1-\sqrt{1+x^{-20}})^{\Large\frac{19}{20}}+2(1-\sqrt{2+x^{-20}})^{\Large-\frac{1}{20}}+C\end{align*}$$