How do you find the indefinite integral of #int 1/(3x+2)#?

Question

Emilia Aguilar · Accepted Answer

See answer below:

Ezekiel Alexander · Answer

undefined To find the indefinite integral of $\int \frac{1}{3x+2} \, dx$, we use the substitution method.

Let $u = 3x + 2$, then $du = 3 \, dx$, or equivalently, $dx = \frac{1}{3} \, du$.

Substituting these expressions into the integral:

$\int \frac{1}{3x+2} \, dx = \int \frac{1}{u} \cdot \frac{1}{3} \, du$

Now, we integrate with respect to $u$:

$\int \frac{1}{u} \cdot \frac{1}{3} \, du = \frac{1}{3} \int \frac{1}{u} \, du$

The integral of $\frac{1}{u}$ with respect to $u$ is $\ln|u| + C$, where $C$ is the constant of integration.

So, substituting back $u = 3x + 2$:

$\frac{1}{3} \int \frac{1}{u} \, du = \frac{1}{3} \ln|3x + 2| + C$

Therefore, the indefinite integral of $\int \frac{1}{3x+2} \, dx$ is $\frac{1}{3} \ln|3x + 2| + C$, where $C$ is the constant of integration.

Paisley Johnson · Answer

undefined To find the indefinite integral of $ \int \frac{1}{3x + 2} $:

1. Recognize that the given function is in the form $ \frac{1}{ax + b} $, which is a standard form for integration.
2. Apply the substitution method:
   Let $ u = 3x + 2 $.
   Then, $ du = 3dx $, or $ dx = \frac{1}{3} du $.
3. Rewrite the integral in terms of $ u $:
   $ \int \frac{1}{u} \cdot \frac{1}{3} du $.
4. Integrate $ \frac{1}{u} $ with respect to $ u $:
   $ \frac{1}{3} \ln|u| + C $.
5. Substitute back $ u = 3x + 2 $ into the result:
   $ \frac{1}{3} \ln|3x + 2| + C $.

So, the indefinite integral of $ \frac{1}{3x + 2} $ is $ \frac{1}{3} \ln|3x + 2| + C $, where $ C $ is the constant of integration.