Integration U substitution then square it I think.

[Zsmitty3]

1. ∫ 1/(2√(x+3)+x)



2. Not sure if I'm beginging this correctly or not but I get stuck.



3. Let u= √x+3 then u² = x+3 2udu=dx dx=2√[(x+3)

Therefore: ∫1/u²-3 Not sure where to go from here?

 

[mtayab1994]

Ok what you did is good but take x from your u^2 you will get x=u^2-3. Then substitute it back into you integral and it should be simple from there.

 

[Zsmitty3]

So plug it into the original integral and get ∫1/(2√(u²-3+3) ?
So the 3's equal 0

Gives you: ∫1(2√u²)

Sqrt and ² cancel leaving you 1/2u

but then that gives you 2u^-1

integrating that gives you u⁰ which is where I keep getting stuck.

 

[Zsmitty3]

If I sub the other way and put x in for u^2-3 then that gives me 1/x. Integral of that is ln absvalue x. When I plug in numbers to make it a definite integral and check it on my calc. that doesn't work.

 

[Curious3141]

but then that gives you 2u^-1

integrating that gives you u⁰ which is where I keep getting stuck.

I didn't read the rest of this, but this is wrong. $\int \frac{1}{u}du = \ln |u| + c$ (c is the constant).

The rule $\int u^n du = \frac{1}{n+1}u^{n+1} + c$ only applies for $n \neq -1$.

 

[Zsmitty3]

So you can pull out the 1/2 and make it .5∫(1/u)

The integral of that is .5(ln absvalue(u))

Sub u back in and get

.5(ln(√x+3)) abs value of √x+3 of course

that's still not giving me a right answer though when I check it.

 

[SammyS]

So plug it into the original integral and get ∫1/(2√(u²-3+3) ?
So the 3's equal 0

Gives you: ∫1(2√u²)

Sqrt and ² cancel leaving you 1/2u

but then that gives you 2u^-1

integrating that gives you u⁰ which is where I keep getting stuck.

No, you don't get 2u^-1 .

\displaystyle \frac{1}{2\sqrt{x+3}+x}\ becomes \displaystyle \ \frac{1}{2u+u^2-3}\ .

 

[Zsmitty3]

-1/u+ln(2u)-u/3 Just plug in √x+3 back in for u.


So that gives you -(1/√x+3)+ln(2√x+3)-(√x+3)/3) ?

 

[SammyS]

-1/u+ln(2u)-u/3 Just plug in √x+3 back in for u.
...

It's not clear how you got that.

The integral \displaystyle \ \int \frac{dx}{2\sqrt{x+3}+x}\ becomes \displaystyle \ \int \frac{2u\,du}{u^2+2u-3}\ .

Now use partial fractions.

 

[Zsmitty3]

You mean partial fraction decomposition or just split them up into separate fractions?

 

[SammyS]

You mean partial fraction decomposition or just split them up into separate fractions?

Yes, partial fraction decomposition. How else can you split them up into separate fractions?

 

[Zsmitty3]

Can you make the u²+2u-3 = (u+3)(u-1)

Then A/(u+3) + B/(u-1) ?

 

[SammyS]

Can you make the u²+2u-3 = (u+3)(u-1)

Then A/(u+3) + B/(u-1) ?

Try it !

 

[Zsmitty3]

So 2=A(u-1)+B(u+3)

Let u= 1

2=A(0)+B(4)

2=3B so B=(1/2)

Let u=-3 and A=-1/2

When you put those in can you just bring the 1/2's out in front of the fraction or do you have to distribute then through the bottom of the integral?

 

[Zsmitty3]

-1/2∫1/(u+3)+1/2∫1/(u-1)

ends up being 1/2* ln(√x+3)-1/√x+3)+3

 

[SammyS]

So 2=A(u-1)+B(u+3)

Let u= 1

2=A(0)+B(4)

2=3B so B=(1/2)

Let u=-3 and A=-1/2

When you put those in can you just bring the 1/2's out in front of the fraction or do you have to distribute then through the bottom of the integral?

Try to be more careful.

That should be:

2u = A(u-1) + B(u+3) .