# Problem Of The Week #418 May 25th, 2020

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#### anemone

##### MHB POTW Director
Staff member
Here is this week's POTW:

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

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#### anemone

##### MHB POTW Director
Staff member
No one answered last week's POTW. However, you can find the suggested solution of other below:

\displaystyle \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

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

\displaystyle \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*}

• • Theia, topsquark, MarkFL and 3 others
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