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integral of x/(x^4 + 25)
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\[u=x^2+25\implies du=?\]
2x dx, so dx = 1/2 du
1/2 integral of du/u
xdx=1/2du
yes
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wait I typed it wrong sorry, it's x^4
\[\int\frac{xdx}{x^4+25}\]?
yes, do I somehow use the derivative of arctan?
yeah that should work
u=x^2
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but derivative of arctan is 1/(x^2 + 1), so how do I get rid of the 25?
\[\int\frac{dx}{x^2+a^2}=\frac1a\tan^{-1}(\frac xa)+C\]if we want to prove that it will take a minute
1/5 arctan (x^2/5)?
careful, you have to make the u-sub first
very close
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1/10 not 1/5 since 1/2 du = dx?
exactly, and don't forget +C if it's indefinite
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