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How to solve for:

$\int \frac{dx}{\left(x^2+4\right)^2}$

What I have tried so far:

Let $x = 2 \tan \theta$
so that $dx = 2 \sec^2\theta d\theta$

$I = \int \frac{2 \sec^2 \theta d\theta}{\left(16 \sec^4\theta \right)^2}$

Not sure where to go from here.

The answer that I expect:

$C + \frac{1}{16} tan^{^-1} \frac{x}{2}+\frac{1}{8}x(x^2+4)^{^-1}$

Meilton
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  • To start off, $$\int \frac1{(x^2+4)^2},\mathrm dx=\frac1{16}\int \frac1{(\frac{x^2}4+1)^2},\mathrm dx$$ now substitute $y=x/2$ to get the above integral – Maximilian Janisch Dec 24 '19 at 15:26
  • Also learn that trigonometric substitutions are useless, except for a handful of hand-crafted problems, which all of them can be solved by other methods. Calculus courses that are not going to teach the better methods should really stop programming students to have trigonometric substitutions as they go-to method. In the linked question, skip the selected answer. That one doesn't teach you anything useful. This answer is better. – MoonLightSyzygy Dec 24 '19 at 15:26
  • @MoonLightSyzygy What would be your favorite method for the above integral? – Maximilian Janisch Dec 24 '19 at 15:27
  • @MaximilianJanisch It is not whether a method is my favourite for a particular integral. What I am saying is that trigonometric substitutions seem to always be taught, in detriment of not teaching the methods that don't fail as often. This creates hordes of students that are easily defeated by integrals that are just small variations of integrals that they have seen solved with trigonometric substitutions. – MoonLightSyzygy Dec 24 '19 at 15:30
  • How did you get $8\color{red}1$? – J. W. Tanner Dec 24 '19 at 15:36

3 Answers3

1

I suggest the following $$I_n=\int\frac{dx}{(x^2+a^2)^n}=\frac{x}{(x^2+a^2)^n}+2n\int\frac{x^2}{(x^2+a^2)^{n+1}}dx.$$ Further we have $\int\frac{x^2}{(x^2+a^2)^{n+1}}dx=I_n-a^2I_{n+1}$, where do we express $I_{n+1}$ through $I_n$.

thing
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Hint:

A basic formula is $$\int\frac{\mathrm dx}{x^2+a^2}=\frac1a\arctan\Bigl(\frac xa\Bigr)$$

On the other hand, if $a\ne 0$, all integrals $$I_n=\int\frac{\mathrm dx}{(x^2+a^2)^n}$$ can be computed recursively, setting $$\begin{cases} u=\dfrac 1{(x^2+a^2)^n} \cr \strut \mathrm dv=\mathrm dx, \end{cases}\quad\text{whence }\;\begin{cases} \mathrm d u=-\dfrac{2nx\mathrm dx}{(x^2+a^2)^{n+1}}\cr u=x \end{cases} $$ and the recurrence relation $$I_{n+1}=\frac{x}{2na^2(x^2+a^2)^n}+\frac{2n-1}{2n}I_n.$$

Bernard
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Even though there's a general method to solve these kinds of integrals (see previous answers) and it's good to know this method, substituting $x=2\tan\theta$ is fine as long as the degree in the denominator is not too large. Note you've made a mistake: $$(x^2+4)^2=(4\tan^2\theta+4)^2=16(1+\tan^2\theta)^2=16\sec^4\theta $$ and the integral becomes $$\int \frac{2\sec^2\theta}{16\sec^4\theta}\mathrm{d}\theta=\frac 18\int \frac{1}{\sec^2\theta}\mathrm{d}\theta=\frac 18\int \cos^2\theta\,\mathrm{d}\theta=\frac 1{16}\int (1+\cos2\theta)\,\mathrm{d}\theta $$ After you solve the integral, you may want to use $$\sin2\theta=\frac{2\tan\theta}{1+\tan^2\theta} $$

J. W. Tanner
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bjorn93
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