Question

How do you find `\int ( \frac { 1} { 3x + 1} ) d x`?

Answer 1

HI,
Lets call `int 1/(3x+1)` as "L"
L=`int (1/(3x+1))(dx)`
We will add and subtract 3x in numerator .

`L=int(3x+1-3x)/(3x+1)(dx)`
now,
`L=int (3x+1)/(3x+1)(dx)-(3x)/(3x+1)(dx)`
`L=int1(dx)-(3x)/(3x+1)(dx)`
As L is integrtion of two integrals
Lets take ,
X=`int1(dx)`
Y=`int(3x)/(3x+1)(dx)`
`L=X-Y`

X can be easily solved
X=`int(x)(dx)`
X=x+C (C is constant)
Lets solve Y

Now lets take

3x=t

So, diffrentiate
`3(dx)=(dt)`
Substitute "dx" in terms of "dt" in Y
Y=`int(t)/(t+1)(dt)/3`
Y=`int(t+1-1)/(t+1)(dt)/3`
Y=`int(t+1)(dt)/3/(t+1)-1/(t+1)(dt)/3`
Y=`t/3-int1/(t+1)*(1/3)`
For this integration you must know formula
`int(f'(x))/f(x)=ln(x)+C`
As diffrentialtion of (t+1) is 1 this`int1/(t+1)` is in this form
Y=`t/3-ln(t+1)/3`+C

Y=`t/3-ln(t+1)/3`+C
3x=t

hence
Y=`x-ln(3x+1)/3`
AS we know
L=X-Y
L=`x-x-ln(3x+1)/3+C`+C+C
C+C is Another constant C
L=`-ln(3x+1)/3`+C

Answer 2

The answer is `=1/3ln(|3x+1|)+C`

Explanation:

We need

`int1/xdx=ln(|x|)+C`

Here,

We perform the substitution

`u=3x+1`, `=>`, `du=3dx`, `dx=1/3du`

Therefore,

`int1/(3x+1)dx=1/3int1/udu=1/3lnu`

`=1/3ln(|3x+1|)+C`

Answer 3

`1/3ln|(3x+1)|+C`

Explanation:

In general

`color(blue)(int(f'(x))/f(x)=ln|f(x)|+C)`

here we have

`int(1/(3x+1))dx`

we note that the top isn't quite the bottom differentiated so let us make an adjustment

`int(1/(3x+1))dx=1/3int 3/(3x+1)dx`

so the numerator is now the denominator differentiated

so we integrate using the above relationship

`=1/3ln|(3x+1)|+C`