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

3 Answers

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 thisint1/(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

Sep 12, 2017

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

Sep 12, 2017

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