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Mathematics 8 Online
OpenStudy (anonymous):

Integration problem.

OpenStudy (unklerhaukus):

yay!

OpenStudy (anonymous):

Do I need to use partial fractions here?

OpenStudy (anonymous):

Don't worry Unkle, my questions on ODEs will end in a 2 week times, stupid repeat exam....

OpenStudy (unklerhaukus):

dam

OpenStudy (athe):

use \[z dz = \frac{1}{2}dz^2\]

OpenStudy (unklerhaukus):

\[\int\frac{z}{z^2+2}\text dz\] ok dont use partial fractions here, remember this integral \[\int\frac{\text dx}x=\ln z+c\]

OpenStudy (unklerhaukus):

let \[x=z^2+2\]\[\text dx=...\text dz\]

OpenStudy (anonymous):

So you are using substitution?

OpenStudy (unklerhaukus):

you could say that

OpenStudy (unklerhaukus):

have you found dx?

OpenStudy (anonymous):

2z?

OpenStudy (unklerhaukus):

2zdz

OpenStudy (anonymous):

So you int I/z dz

OpenStudy (anonymous):

ln z +c

OpenStudy (unklerhaukus):

\[\frac12\int\frac{2z\cdot dz}{z^2+2}=\qquad\frac 12\int\frac{dx}{x}=\frac 12\ln x+c\] \[=\frac12\ln (z^2+2)+c\]

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