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Mathematics 17 Online
Parth (parthkohli):

\[\int {3 \over 5y + 4}dy\]

Parth (parthkohli):

I'd first take \(u = 5y + 4\), then \(du = 5dy\)?

OpenStudy (anonymous):

yup

Parth (parthkohli):

So,\[dy = {1 \over 5}du\]

OpenStudy (anonymous):

yeah, then you are almost done!

Parth (parthkohli):

OK, let's make the substitutions.\[{1 \over 5}\int{3 \over u}du \]

OpenStudy (anonymous):

you can also take the 3 out of the integral

Parth (parthkohli):

Yes, doing that.\[{3 \over 5}\int{1 \over u} du \]

Parth (parthkohli):

So ln(u) inside?

Parth (parthkohli):

\[{3 \over 5} \ln(u) + c \]Is that correct?

OpenStudy (turingtest):

now sub back in for u

Parth (parthkohli):

And then substituting back,\[ {3 \over 5}\ln(5y +4) + c\]

Parth (parthkohli):

Right?

OpenStudy (turingtest):

yessir :)

Parth (parthkohli):

Thank you :)

hartnn (hartnn):

more accurately its 3/5 ln |5y+4| + c

Parth (parthkohli):

Oh... I forgot the |...| brackets.

hartnn (hartnn):

those brackets should be there with ln everytime.

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