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Integration problem.
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yay!
Do I need to use partial fractions here?
Don't worry Unkle, my questions on ODEs will end in a 2 week times, stupid repeat exam....
dam
use \[z dz = \frac{1}{2}dz^2\]
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\[\int\frac{z}{z^2+2}\text dz\] ok dont use partial fractions here, remember this integral \[\int\frac{\text dx}x=\ln z+c\]
let \[x=z^2+2\]\[\text dx=...\text dz\]
So you are using substitution?
you could say that
have you found dx?
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2z?
2zdz
So you int I/z dz
ln z +c
\[\frac12\int\frac{2z\cdot dz}{z^2+2}=\qquad\frac 12\int\frac{dx}{x}=\frac 12\ln x+c\] \[=\frac12\ln (z^2+2)+c\]
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