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\[\int {3 \over 5y + 4}dy\]
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I'd first take \(u = 5y + 4\), then \(du = 5dy\)?
yup
So,\[dy = {1 \over 5}du\]
yeah, then you are almost done!
OK, let's make the substitutions.\[{1 \over 5}\int{3 \over u}du \]
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you can also take the 3 out of the integral
Yes, doing that.\[{3 \over 5}\int{1 \over u} du \]
So ln(u) inside?
\[{3 \over 5} \ln(u) + c \]Is that correct?
now sub back in for u
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And then substituting back,\[ {3 \over 5}\ln(5y +4) + c\]
Right?
yessir :)
Thank you :)
more accurately its 3/5 ln |5y+4| + c
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Oh... I forgot the |...| brackets.
those brackets should be there with ln everytime.
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