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Let $n>1$ be an integer and $a \neq 0$ a constant. Prove the reduction formula $$\int \frac{1}{(x^2+a^2)^n} \, dx = \frac{1}{2a^2(n-1)} \left[ \frac{x}{(x^2+a^2)^{n-1}} + (2n-3) \int \frac{1}{(x^2+a^2)^{n-1}} \, dx \right].$$ Wanting to prove this ... Having difficulty knowing where to start/how to proceed.

Mark Fantini
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Guest
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    Applying the trigonometric substitution $ \ x \ = \ a \tan \theta \ , \ dx \ = \ a \sec^2 \theta \ d\theta \ $ , the integral becomes $$ \rightarrow \ \ \frac{1}{a^{2n-1}} \int \ \cos^{2n-2} \theta \ \ d\theta \ \ , $$ which is easier to deal with. – colormegone May 11 '14 at 06:14
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    Reduction formula 'always' involves IBP. – Tunk-Fey May 11 '14 at 06:42

2 Answers2

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We proceed by integration by parts. Assume $n>1$ so that $n-1>0$ then we have:

$$\int\frac{1}{(x^{2}+a^{2})^{n-1}}dx=\frac{x}{(x^{2}+a^{2})^{n-1}}+2(n-1)\int\frac{x^{2}}{(x^{2}+a^{2})^{n}}dx$$

$$=\frac{x}{(x^{2}+a^{2})^{n-1}}+2(n-1)\int\frac{1}{(x^{2}+a^{2})^{n-1}}dx-2(n-1)a^{2}\int\frac{1}{(x^{2}+a^{2})^{n}}dx$$

Now rearranging the terms we get:

$$\int\frac{1}{(x^{2}+a^{2})^{n}}dx=\frac{1}{2(n-1)a^{2}}\bigg(\frac{x}{(x^{2}+a^{2})^{n-1}}+(2n-3)\int\frac{1}{(x^{2}+a^{2})^{n}}dx\bigg)$$

user71352
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  • How come you start with the exponent n-1 ? should it not be n?

    Also, what substitution did you use (for u and dv)?

     – Guest May 12 '14 at 17:50
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    I started with $n-1$ because I wanted my resulting formula to match yours. Starting the exponent with $n$ would give the same information. When I integrated by parts I used $dv=1$, $v=x$, $du=\frac{2(n-1)x}{(x^{2}+a^{2})^{n}}$, and $u=\frac{1}{(x^{2}+a^{2})^{n-1}}$. – user71352 May 13 '14 at 05:43
  • Thanks for the reply. Just 1 more question ... Could you explain how you got from the first line to the second line.

    What I mean is that from the first line where did the integral(x^2 / ( x^2 + A^2)^n)dx go?

     – Guest May 15 '14 at 01:38
  • You're welcome. I used that $\int\frac{x^{2}}{(x^{2}+a^{2})^{n}}dx=\int\frac{(x^{2}+a^{2})-a^{2}}{(x^{2}+a^{2})^{n}}dx=\int\frac{1}{(x^{2}+a^{2})^{n-1}}dx-\int\frac{a^{2}}{(x^{2}+a^{2})^{n}}dx$. – user71352 May 15 '14 at 05:21
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Hint: you can always prove an indefinite integration formula by differentiating both sides.

Robert Israel
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