evaluate integral problem?

Question

hi. i would like to find out how to solve this problem. I asked this question before also and deleted it as no one could come up with the right answer at the time and very few people could answer the question correctly

anyway....

the question i wish to solve is i would like to evaluate the integral

∫ln( √3x) dx

thanks for your help and time with this. it ismuch appreciated

the seond one is right

Answer 1

I'm not sure of precisely what you're asking; there are two possibilities, depending on whether the x is supposed to be inside the square root or not. I'll answer both below.

Note that ∫ln x dx
= ∫ln x (1 dx)
= x ln x - ∫x(1/x) dx
= x ln x - x + c.

∫ln ((√3) x) dx
= ∫(ln x + (1/2) ln 3) dx
= x ln x - x + (x ln 3) / 2 + c

∫ln √(3x) dx
= ∫(1/2) ln (3x) dx
= (1/2) [(3x) ln (3x) - (3x)] / 3 + c
= (x ln (3x) - x) / 2 + c.

Answer 2

K I don't know how to do the integral sign because I just got a mac and I don't even know how to get into like a word-like program to find the symbol so I'll just write 'integral' k? And I promise, I'm better at math than computers...

so the answer is: 1/2 (xln3x-(1/3)x) + c

The solution:
integral ln(root3x)dx
= integral ln 3x^1/2 dx
= integral 1/2 ln 3x dx
= 1/2 integral ln 3x dx
Then do integration by parts...
= 1/2 (xln3x - integral (x/3x) dx)
= 1/2 (xln3x - 1/3x) + c

Answer 3

Sit down, strip in, buckle up and get ready for a supersonic blast. By the end of this you won't know what hit you.

∫ln( √3x) dx
∫ln(3x)^½ dx
∫½ ln(3x) dx
∫½ (ln 3 + ln x) dx
½ [ x ln 3 + x ln x - x) ] + c
½ [ x (ln 3 + ln x -1) ] + c

Now, let's differentiate the answer to see if the answer is correct.

dy/dx = ½ [ x (ln 3 + ln x -1) ] + c dx
dy/dx = ½ [ x ln 3 + x ln x - x) ] + c dx
dy/dx = ½ [ ln 3 + x(1/x) + ln x(1) - 1] + 0
dy/dx = ½ [ ln 3 + (1) + ln x - 1]
dy/dx = ½ [ ln 3 + ln x ]
dy/dx = ½ [ ln 3x ]
dy/dx = ln (3x)^½
dy/dx = ln √(3x)


Or if you mean:
∫ln √3(x) dx
∫[ ln√3 + ln(x)] dx
x ln√3 + x ln(x) - x + c
x [ ln√3 + ln(x) - 1 ] + c
x [ ln√3(x) - 1 ] + c
Differentiate the answer to see if the answer is correct.
x [ ln√3(x) - 1 ] + c
x [ ln√3 + ln(x) - 1 ] + c
x ln√3 + x ln(x) - x + c
dy/dx = x ln√3 + x ln(x) - x + c dx
dy/dx = ln√3 + x(1/x) + (ln x)(1) - 1 + 0
dy/dx = ln√3 + (1) + ln x - 1
dy/dx = ln√3 + ln x
dy/dx = ln(√3)(x)


It's as simple as that.

Answer 4

You can simplify this problem by using the laws of logs

ln( √3x) = ln√3 + lnx

∫(ln√3x) dx = ∫ln√3dx + ∫lnx dx

..................= xln√3 + xlnx - lnx + c

or if you meant

ln( √3 √x) = ln√3 + ln√x = ln√3 + ½ lnx

∫(ln√3x) dx = ∫ln√3dx + ½∫(lnx) dx

.................. = xln√3 + ½(xlnx - lnx) + c

Answer 5

Complete squares (x^2-2x+2)= ((x^2-2x+1) +1) = ( x-1)^2+1 INT 8 dx /( x-1)^2+1 (x-1) =Tan u dx= sec^2u du (x-1)^2+1 = sec^2 u INT 8du = 8u = 8Arctan (x-1) 1

Answer 6

∫ln(x)dx
=∫ln(x).1dx

Integration by parts
∫udv = uv-∫vdu

u=ln(x)
du=(1/x)dx

dv=1
v=x

∫ln(x)dx
= xln(x)- ∫x.(1/x)dx
= xln(x) - x+c


∫ln( √3x) dx
= ∫ln(√3)+ ln(x)dx
= xln(√3) + ∫ln(x)dx
= xln(√3) + xln(x) - x + c

If you meant
∫ln( √(3x)) dx
Then this equals
= ∫ln(√3)+ ln(√x)dx
= xln(√3) + (1/2)∫ln(x)dx
= xln(√3) + (x/2)ln(x) - x/2 + c