Welcome to our community

Be a part of something great, join today!

Problem Of The Week #418 May 25th, 2020

Status
Not open for further replies.
  • Thread starter
  • Admin
  • #1

anemone

MHB POTW Director
Staff member
Feb 14, 2012
3,586
Here is this week's POTW:

-----

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

-----

Remember to read the POTW submission guidelines to find out how to submit your answers!
 
  • Thread starter
  • Admin
  • #2

anemone

MHB POTW Director
Staff member
Feb 14, 2012
3,586
No one answered last week's POTW. (Sadface) 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

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

\(\displaystyle \begin{align*} \therefore \int \left(x^{10}+\sqrt{1+x^{20}}\right)^{^{\Large\frac{21}{10}}}\,dx&=\int u^{\Large-\frac{21}{10}}\left(-\dfrac{1}{10}(u+1)\right)\,du\\&=-\dfrac{1}{10} \int (u^{\Large-\frac{1}{20}}+u^{\Large-\frac{21}{20}})\,du\\&=-\dfrac{1}{10}\left(\dfrac{20}{19}u^{\Large\frac{19}{20}}-20u^{\Large-\frac{1}{20}}\right)+C\\&=-\dfrac{2}{19}(1-\sqrt{1+x^{-20}})^{\Large\frac{19}{20}}+2(1-\sqrt{2+x^{-20}})^{\Large-\frac{1}{20}}+C\end{align*}\)
 
Status
Not open for further replies.