What is the integral of (x+2)/sqrt (3x-1)?

[polepole]

$$\int$$$$(X + 2)/\sqrt{3X-1}$$




one of my attempts: u=3X-1 , du/3=dX , x = (u+1)/3

1/3$$\int$$$$(u+7).du/\sqrt{u}$$ = 1/3$$\int$$$$u .du/\sqrt{u}$$ + 7/3 $$\int$$$$du/\sqrt{u}$$ = 1/3$$\int$$$$u .du/\sqrt{u}$$ + 14/3. sqrt{u} = stuck...

the answer should be 2/27 . (3X + 20) .$$\sqrt{3X-1}$$

 

[hgfalling]

I'm not sure I'm a hundred percent following your work, but it seems that you are just making algebra errors.

The substitution you are intending to make is fine.

Then you write:

$$\frac{1}{3} \int \frac {(u+7)}{\sqrt{u}} du$$

as the integral after substitution. But this isn't right; plug u back in here, you'll find that the coefficient at front should be something different. Then you seem to have taken that 7 out as though it were a constant, but you can't do that either. Anyway, you are on the right track with substituting u = 3x - 1. Just carefully redo the algebra.

 

[polepole]

I forgot to multiply with 1/3 from du/3. becomes:

1/9$$\int$$$$(u+7).du/\sqrt{u}$$ = 1/9$$\int$$$$u .du/\sqrt{u}$$ + 7/9 $$\int$$$$du/\sqrt{u}$$ = 1/9$$\int$$$$u .du/\sqrt{u}$$ + $$14/3. \int \sqrt{u}$$

but I've found the solution from there.
btw: if you split the integral you can use 7 as an a constant and put it in front of the intgr sign

 

[hgfalling]

Oh, I see what you did. Yes, the 7 was fine, just the multiplier was off.

 

[Dickfore]

Integrals of the type:

$$\int{R(x, \sqrt[n]{a x + b}) \, dx}$$

where $R(u, v)$ is a rational function of both of its arguments can always be reduced to integrals of rational functions with the substitution:

$$t = \sqrt[n]{a x + b}$$

 

[polepole]

So you would've used t=sqrt(3X-1) ?

 

[Dickfore]

So you would've used t=sqrt(3X-1) ?

Yes.