$$ \int \frac{dx}{\left(x^2+9\right)^2}$$
How would you solve this with partial integration (without trigonometry)?
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What have you tried ? – François Mortier Mar 30 '24 at 16:31
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https://math.stackexchange.com/a/689932/1242 – Hans Lundmark Mar 30 '24 at 16:34
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Do you mean "partial fractions?" – Thomas Andrews Mar 30 '24 at 16:35
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1@FrançoisMortier I tried something like this but I don't know what to do next
– Nil
Mar 30 '24 at 16:40
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@HansLundmark thank you – Nil Mar 30 '24 at 16:41
1 Answers
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$$I=\int 1.\frac{1}{x^2+9}dx=x.\frac{1}{x^2+9}-\int x\left(-\frac{1}{(x^2+9)^2}\right)(2x)dx$$$$=\frac{x}{x^2+9}+2\int \frac{x^2}{(x^2+9)^2}dx$$
$$I=\frac{x}{x^2+9}+2\int \frac{x^2+9-9}{(x^2+9)^2}dx=\frac{x}{x^2+9}+2\int \frac{1}{(x^2+9)}dx-18\int \frac{1}{(x^2+9)^2}dx$$$$I=\frac{x}{x^2+9}+2I-18\int \frac{1}{(x^2+9)^2}dx$$
$$18\int \frac{1}{(x^2+9)^2}dx=\frac{x}{x^2+9}+I=\frac{x}{x^2+9}+\frac{1}{3}\tan^{-1}\frac{x}{3}+C$$
$$\int \frac{1}{(x^2+9)^2}dx=\frac{x}{x^2+9}+I=\frac{x}{18(x^2+9)}+\frac{1}{54}\tan^{-1}\frac{x}{3}+C$$
Maverick
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