does squaring a rational expression inside an integral affect dx?

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does squaring a rational expression inside an integral affect dx?

Mathematics
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|dw:1433407624358:dw|I plan to do it on a problem that looks like this
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Try something like \[I=- \int\limits \frac{-x^2dx}{\sqrt{-x^2-4x+5}}\]\[I=-(\int\limits \frac{(-x^2-4x+5+4x-5)dx}{\sqrt{-x^2-4x+5}})\]\[I=-(\int\limits \frac{(-x^2-4x+5)dx}{\sqrt{-x^2-4x+5}}+4\int\limits \frac{xdx}{\sqrt{-x^2-4x+5}}-5\int\limits \frac{dx}{\sqrt{-x^2-4x+5}})\]

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Other answers:

First integral factorize in the form of \[a^2-(x+z)^2\] substitute x+z so it becomes \[a^2-t^2\] Use the formula \[\int\limits \sqrt{a^2-t^2} dt=\frac{t \sqrt{a^2-t^2}}{2}+\frac{a^2}{2}\sin^{-1}(\frac{t}{a})\] Second integral Let \[x=A \frac{d(-x^2-4x+5)}{dx}+B\] Equate coefficients and solve Third Integral Factorize and apply \[I=\int\limits \frac{dt}{\sqrt{a^2-t^2}}=\sin^{-1}\frac{t}{a}\]
So sorry, I was solving other problems and skipped this one. Thank you so much

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