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

HI,
Lets call $\int \frac{1}{3 x + 1}$ as "L"
L=$\int \left(\frac{1}{3 x + 1}\right) \left(\mathrm{dx}\right)$
We will add and subtract 3x in numerator .

$L = \int \frac{3 x + 1 - 3 x}{3 x + 1} \left(\mathrm{dx}\right)$
now,
$L = \int \frac{3 x + 1}{3 x + 1} \left(\mathrm{dx}\right) - \frac{3 x}{3 x + 1} \left(\mathrm{dx}\right)$
$L = \int 1 \left(\mathrm{dx}\right) - \frac{3 x}{3 x + 1} \left(\mathrm{dx}\right)$
As L is integrtion of two integrals
Lets take ,
X=$\int 1 \left(\mathrm{dx}\right)$
Y=$\int \frac{3 x}{3 x + 1} \left(\mathrm{dx}\right)$
$L = X - Y$

X can be easily solved
X=$\int \left(x\right) \left(\mathrm{dx}\right)$
X=x+C (C is constant)
Lets solve Y

Now lets take

3x=t

So, diffrentiate
$3 \left(\mathrm{dx}\right) = \left(\mathrm{dt}\right)$
Substitute "dx" in terms of "dt" in Y
Y=$\int \frac{t}{t + 1} \frac{\mathrm{dt}}{3}$
Y=$\int \frac{t + 1 - 1}{t + 1} \frac{\mathrm{dt}}{3}$
Y=$\int \left(t + 1\right) \frac{\mathrm{dt}}{3} / \left(t + 1\right) - \frac{1}{t + 1} \frac{\mathrm{dt}}{3}$
Y=$\frac{t}{3} - \int \frac{1}{t + 1} \cdot \left(\frac{1}{3}\right)$
For this integration you must know formula
$\int \frac{f ' \left(x\right)}{f} \left(x\right) = \ln \left(x\right) + C$
As diffrentialtion of (t+1) is 1 this$\int \frac{1}{t + 1}$ is in this form
Y=$\frac{t}{3} - \ln \frac{t + 1}{3}$+C

Y=$\frac{t}{3} - \ln \frac{t + 1}{3}$+C
3x=t

hence
Y=$x - \ln \frac{3 x + 1}{3}$
AS we know
L=X-Y
L=$x - x - \ln \frac{3 x + 1}{3} + C$+C+C
C+C is Another constant C
L=$- \ln \frac{3 x + 1}{3}$+C

Sep 12, 2017

The answer is $= \frac{1}{3} \ln \left(| 3 x + 1 |\right) + C$

#### Explanation:

We need

$\int \frac{1}{x} \mathrm{dx} = \ln \left(| x |\right) + C$

Here,

We perform the substitution

$u = 3 x + 1$, $\implies$, $\mathrm{du} = 3 \mathrm{dx}$, $\mathrm{dx} = \frac{1}{3} \mathrm{du}$

Therefore,

$\int \frac{1}{3 x + 1} \mathrm{dx} = \frac{1}{3} \int \frac{1}{u} \mathrm{du} = \frac{1}{3} \ln u$

$= \frac{1}{3} \ln \left(| 3 x + 1 |\right) + C$

Sep 12, 2017

$\frac{1}{3} \ln | \left(3 x + 1\right) | + C$

#### Explanation:

In general

$\textcolor{b l u e}{\int \frac{f ' \left(x\right)}{f} \left(x\right) = \ln | f \left(x\right) | + C}$

here we have

$\int \left(\frac{1}{3 x + 1}\right) \mathrm{dx}$

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

$\int \left(\frac{1}{3 x + 1}\right) \mathrm{dx} = \frac{1}{3} \int \frac{3}{3 x + 1} \mathrm{dx}$

so the numerator is now the denominator differentiated

so we integrate using the above relationship

$= \frac{1}{3} \ln | \left(3 x + 1\right) | + C$