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So I have this integral $\displaystyle\int \frac{1}{1+x^2\cdot y^2}dy$. Is it possible to rewrite this integral to $\displaystyle\int \frac{1}{1+u^2}du$ using substitution, as I know the value of this integral, that is I want to remove $x^2$ from the denominator somehow. I'm not sure how.

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  • $\begingroup$ Btw, the limits on these integrals are $-\infty$ and $\infty$. $\endgroup$ – seht111 Jan 19 '17 at 17:40
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    $\begingroup$ Since you are only integrating with respect to $y$ you may treat $x^2$ like a constant. When you realize this it is easy to recognize this integral as a standard one. $\endgroup$ – Vik78 Jan 19 '17 at 17:41
  • $\begingroup$ @seht111, then you should write $\int_{-\infty}^\infty$ instead of just $\int$. You can still edit you post $\endgroup$ – Yuriy S Jan 19 '17 at 21:24
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Setting $$xy=t$$ then we get $$xdy=dt$$ and our integral will be $$\frac{1}{x}\int\frac{1}{1+t^2}dt$$

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\begin{align*} \int \frac{1}{1+x^2 y^2} \, dy &= \frac1{x^2} \int \frac{1}{\frac1{x^2}+y^2} \, dy \\ &= \frac1{x^2} \int \frac{1}{\left( \frac1{x} \right)^2+y^2} \, dy \\ &= \frac1{x^2} \cdot \frac1{\frac1{x}} \cdot \tan^{-1} \frac{y}{\frac1{x}} + C \\ &= \frac1{x} \tan^{-1} xy + C \end{align*}

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