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Your question \[\int\limits \frac{7}{\sqrt{5+4x-x^2}}\] Try completing the square @tanvidais13
wait i don't need to, its factorizable: \[7/\sqrt{(1+x)(5-x)}\]
no you can't do that because we can't use partial fractions here

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so we need to complete the square, or do a substitution?
We can try using trig substitution, it might work, so if we complete the square then we got \[\int\limits\limits \frac{7}{\sqrt{9-(x-2)^2}} \, dx\] Put it like this, \[\int\limits\limits\limits \frac{7}{\sqrt{3^2-(x-2)^2}} \, dx\] Using trig sub, \[x=asin(\theta), \text{while} ~~~~~ a^2-x^2 ~~~~~ \text{is} ~~~~~ 3^2-(x-2)^2, \\ \\So \\ \\------------\\\\ x-2=3\sin(\theta) ~~~~~ \theta=\sin^{-1}(\frac{x}{3}) \\ \\ dx=3\cos (\theta) d \theta \\\\-----------\\\\ 7 \int\limits \frac{3\cos \theta}{\sqrt{3^2-(3\sin \theta)^2}}d \theta\]
\[7 \int\limits \frac{3\cos \theta}{\sqrt{3^2(1-\sin^2 \theta)}}d \theta \\ \\ \text{We know that} \sqrt{ab}=\sqrt{a} \sqrt{b}\\ \\ 7 \int\limits \frac{\cos \theta}{\sqrt{(1-\sin^2 \theta)}}d \theta \\ \\ 7 \int\limits \frac{\cos \theta}{\sqrt{\cos^2 \theta}}d \theta \\ \\7 \int\limits 1 d \theta\]
\[7 \theta +c \\ \\ 7\sin^{-1}(\frac{x-2}{3}) +c\]
Note that theta is\[\theta=\sin^{-1}(\frac{u}{3})\] and u=x-2
@tanvidais13 Both completing the square and substitution needed
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