# How do you evaluate the integral int ln(3x+4)?

Apr 15, 2018

$\frac{3 x + 4}{3} \left(\ln \left(3 x + 4\right) - 1\right) + C$

#### Explanation:

Let $I = \int \ln \left(3 x + 4\right) \mathrm{dx}$

N.B. Since $I$ is an indefinite integral it cannot be "evaluated". Rather it can be expressed as a function of $x$ plus an arbitrary constant.

Let $u = \left(3 x + 4\right) \to \mathrm{dx} = \frac{\mathrm{du}}{3}$

$\therefore I = \int \frac{\ln u}{3} \mathrm{du} = \frac{1}{3} \int \ln u \mathrm{du}$

Integration by Parts states: $\int f \left(x\right) g ' \left(x\right) \mathrm{dx} = f \left(x\right) g \left(x\right) - \int f ' \left(x\right) g \left(x\right) \mathrm{dx}$

Considering $3 I = \int \ln u \cdot 1 \mathrm{du}$

$f \left(x\right) = \ln u , g ' \left(x\right) = 1$

Thus, $f ' \left(x\right) = \frac{1}{u} , g \left(x\right) = u$

$\therefore 3 I = \ln u \cdot u - \int \left(\frac{1}{u} \cdot u\right) \mathrm{du}$

$= u \ln u - \int 1 \mathrm{du}$

$= u \ln u - u$

$\therefore I = \frac{u}{3} \left(\ln u - 1\right) + C$

Undo substitution.

$I = \frac{3 x + 4}{3} \left(\ln \left(3 x + 4\right) - 1\right) + C$

Apr 15, 2018

$\frac{1}{3} \left(3 x + 4\right) \left(\ln \left(3 x + 4\right) - 1\right) + C$

#### Explanation:

We got: $\int \ln \left(3 x + 4\right) \setminus \mathrm{dx}$

Let's use u-substitution. Let $u = 3 x + 4 , \therefore \mathrm{du} = 3 \setminus \mathrm{dx} , \mathrm{dx} = \frac{\mathrm{du}}{3}$

And so,

$= \int \ln u \setminus \frac{\mathrm{du}}{3}$

Taking out the constant, the problem becomes,

$= \frac{1}{3} \int \ln u \setminus \mathrm{du}$

Let's find $\int \ln u \setminus \mathrm{du}$. We got:

$\int \ln u \setminus \mathrm{du}$

$= \int \ln u \cdot 1 \setminus \mathrm{du}$

Now, we use integration by parts, which is:

$\int u \setminus \mathrm{dv} = u v - \int v \setminus \mathrm{du}$

Let $u = \ln u , \mathrm{dv} = 1$.

$\therefore v = u , \mathrm{du} = \frac{1}{u}$

And so,

$\int \ln u \cdot 1 \setminus \mathrm{du} = u \setminus \ln u - \int u \cdot \frac{1}{u} \setminus \mathrm{du}$

$= u \setminus \ln u - \int 1 \setminus \mathrm{du}$

$= u \setminus \ln u - u + C$

Now, we plug that back into the original integral.

We get:

$= \frac{1}{3} \left(u \setminus \ln u - u\right)$

Notice how I don't put in the constant yet, as we always put the constant after the final calculation, and not during the calculation, as $\frac{1}{3} C$ would be incorrect.

$= \frac{1}{3} u \setminus \ln u - \frac{1}{3} u$

Substitute back $u = 3 x + 4$, we get:

$= \frac{1}{3} \left(3 x + 4\right) \ln \left(3 x + 4\right) - \frac{1}{3} \left(3 x + 4\right)$

$= \frac{1}{3} \left(3 x + 4\right) \left(\ln \left(3 x + 4\right) - 1\right)$

Now, we can add the constant.

$= \frac{1}{3} \left(3 x + 4\right) \left(\ln \left(3 x + 4\right) - 1\right) + C$