Intergral question

[markraz]

Hi again, I have this rather simple looking integral but when I get to a point I'm not sure which way to proceed


\(\displaystyle \Large\int (x+1) \sqrt{3x+2} dx\)

\(\displaystyle \Large\int (x+1) ({3x+2})^{\frac{1}{2}} dx\)

\(\displaystyle \Large u = (3x+2)\)

\(\displaystyle \Large\int (x+1) (u)^{\frac{1}{2}} dx\)


\(\displaystyle \Large du = 3\)
\(\displaystyle \Large dx = \frac{1}{3} du\)

replace DX with DU
\(\displaystyle \Large\int (x+1) (u)^{\frac{1}{2}} dx\)
\(\displaystyle \Large\int (x+1) (u)^{\frac{1}{2}}(\frac{1}{3})du\)


\(\displaystyle \Large x = \frac{u-2}{3} du\)

Now respect to X
\(\displaystyle \Large(\frac{1}{3})\int (x+1) (u)^{\frac{1}{2}}du\)

Now respect to X in first term
\(\displaystyle \Large(\frac{1}{3})\int ({ (\frac{u-2}{3})}+1) (u)^{\frac{1}{2}}du\)

ok here is my question....
which would you do first

1. distribute the \(\displaystyle {\sqrt{u}}\) with \(\displaystyle ({ (\frac{u-2}{3})}+1)\) ?
Or
2. Should I expand out the \(\displaystyle (({\frac{u}{3}}) -{ (\frac{2}{3})} +1) \) then distribute the \(\displaystyle {\sqrt{u}}\) ?
I tried both ways, but they didn't seem like they were going to yield the same result

I realize this is more basic algebra but I figured I'd ask here

Thanks

 

[pka]

Hi again I have this rather simple looking integral but when I get to a point I'm not sure which way to proceed
\(\displaystyle \Large\int (x+1) \sqrt{3x+2} dx\)

It appears to as if you have no idea about any of this.

\(\displaystyle u = 3x + 2,\quad x = \dfrac{{u - 2}}{3},\quad dx = \left( {\frac{1}{3}} \right)du\)

Thus \(\displaystyle \displaystyle\int {\left[ {\left( {\frac{{u - 2}}{3}} \right) + 1} \right]\sqrt u \left( {{3^{ - 1}}du} \right)} \)

 

[markraz]

It appears to as if you have no idea about any of this.

yes sir, hence why I posted a question asking for advice..


anyhoo

Is there anyone here who can help me out?

thanks in advance

 

[Ishuda]

I would say 2 and also pull out the 1/3 to write the whole thing as
\(\displaystyle \frac{1}{9} (u + 1)\space u^{\frac{1}{2}} \)

 

[markraz]

I would say 2 and also pull out the 1/3 to write the whole thing as
\(\displaystyle \frac{1}{9} (u + 1)\space u^{\frac{1}{2}} \)

thanks so much..
so it would look as such?

\(\displaystyle \Large\frac{1}{3}\int [({\frac{u}{3}}) -{ (\frac{2}{3})} +1] (u)^{\frac{1}{2}}\)

then I would distribute the \(\displaystyle (u)^{\frac{1}{2}}\) to each term inside the \(\displaystyle [({\frac{u}{3}}) -{ (\frac{2}{3})} +1]\) ??

thanks again

 

[Ishuda]

thanks so much..
so it would look as such?

\(\displaystyle \Large\frac{1}{3}\int [({\frac{u}{3}}) -{ (\frac{2}{3})} +1] (u)^{\frac{1}{2}}\)

then I would distribute the \(\displaystyle (u)^{\frac{1}{2}}\) to each term inside the \(\displaystyle [({\frac{u}{3}}) -{ (\frac{2}{3})} +1]\) ??

thanks again

Simplify and pull out the 1/3 first, then distribute the \(\displaystyle u^{\frac{1}{2}}\)

 

[markraz]

thanks