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Mathematics 16 Online
OpenStudy (anonymous):

Find the integral!!!

OpenStudy (anonymous):

\[\int\limits_{}^{} \frac{ 3 }{ \sqrt[3]{3x-5} } dx\]

OpenStudy (anonymous):

I really don't have much a clue for this one

OpenStudy (anonymous):

Let 3x - 5 = u

OpenStudy (anonymous):

Now..Use Substitution Method

OpenStudy (anonymous):

and I take out the 3 on top right?

OpenStudy (mimi_x3):

\(u=3x+5\) du/dx=3 \[\int\limits\frac{3}{\sqrt[3]{u}} *\frac{du}{3} =>\int\limits\frac{1}{\sqrt[3]{u}} du=>\int\limits\frac{1}{u^{1/3}} du\]

OpenStudy (anonymous):

du / dx = 3 du = 3 dx

OpenStudy (mimi_x3):

u=3x-5**

OpenStudy (anonymous):

perfect! I'll try it again

OpenStudy (anonymous):

would it be \[\frac{ 3(3x-5)^{2/3} }{ 2 } +c\]

OpenStudy (anonymous):

is this method called integration by substitution method??

OpenStudy (anonymous):

I think so

OpenStudy (sirm3d):

you can also try \[y=\sqrt[3]{3x-5}\Rightarrow y^3=3x-5\\3y^2\;\mathrm dy =3\;\mathrm dx\]

OpenStudy (anonymous):

yup.....also an other way is by doing direct inverse power......

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