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Problem of the Week #44 - January 28th, 2013

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Chris L T521

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Jan 26, 2012
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Thanks to those who participated in last week's POTW!! Here's this week's problem!

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Problem: Let $J_n=\displaystyle\int\frac{\,dx}{(x^2+1)^n}$. Use integration by parts to prove that
\[J_{n+1} = \left(1-\frac{1}{2n}\right)J_n+\left(\frac{1}{2n}\right) \frac{x}{(x^2+1)^n}.\]

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Remember to read the POTW submission guidelines to find out how to submit your answers!
 
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Chris L T521

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Jan 26, 2012
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This weeks question was correctly answered by BAdhi, MarkFL and Sudharaka. You can find BAdhi's solution below:

It is given that,$$J_n=\int \frac{1}{(x^2+1)^n} \, dx$$
If we consider $\int \frac{1}{(x^2+1)^n} \, dx$ and with integral by parts,
$$\begin{align*}
\int \frac{1}{(x^2+1)^n} \, dx &= \int \frac{d\,x}{dx}\cdot \frac{1}{(x^2+1)^n} \, dx\\
&=x\cdot \frac{1}{(x^2+1)^2}-\int x\cdot \left[\frac{-2nx}{(x^2+1)^{(n+1)}}\right]\,dx\\
&=\frac{x}{(x^2+1)^2}+2n\int \frac{x^2}{(x^2+1)^{(n+1)}} \,dx\\
\frac{1}{2n}J_n&=\left(\frac{1}{2n}\right) \frac{x}{(x^2+1)^2}+\int \frac{x^2}{(x^2+1)^{(n+1)}} \,dx
\end{align*}$$
By adding $-J_n$ to both sides,
$$\begin{align*}
-J_n +\frac{1}{2n}J_n&=\left(\frac{1}{2n}\right) \frac{x}{(x^2+1)^2}+\int \frac{x^2}{(x^2+1)^{(n+1)}} \,dx-J_n\\
\left(\frac{1}{2n}-1\right) J_n & = \left(\frac{1}{2n}\right) \frac{x}{(x^2+1)^2}+\int \frac{x^2}{(x^2+1)^{(n+1)}} -\frac{1}{(x^2+1)^n}\,dx\\
&=\left(\frac{1}{2n}\right) \frac{x}{(x^2+1)^2} + \int \frac{x^2-(x^2+1)}{(x^2+1)^{(n+1)}}\,dx\\
&=\left(\frac{1}{2n}\right) \frac{x}{(x^2+1)^2}-\underbrace{\int \frac{1}{(x^2+1)^{(n+1)}}\,dx}_{J_{(n+1)}}\\
&=\left(\frac{1}{2n}\right) \frac{x}{(x^2+1)^2}-J_{(n+1)}
\end{align*}$$
By adjusting,
$$J_{(n+1)}=\left(1-\frac{1}{2n}\right) J_n+\left(\frac{1}{2n}\right) \frac{x}{(x^2+1)^2}$$
 
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