To solve the integral [tex]\(\int \frac{1}{2x + 1} \, dx\)[/tex], we will follow these steps:
1. Identify the integrand and recognize its form:
We need to find [tex]\(\int \frac{1}{2x + 1} \, dx\)[/tex].
2. Substitute and simplify if necessary:
Notice that the integrand [tex]\(\frac{1}{2x + 1}\)[/tex] has a linear function in the denominator. A common technique for such integrals is to let [tex]\(u = 2x + 1\)[/tex]. This substitution simplifies the integrand.
Let [tex]\( u = 2x + 1 \)[/tex].
3. Differentiate [tex]\( u \)[/tex] with respect to [tex]\( x \)[/tex]:
[tex]\[
\frac{du}{dx} = 2 \implies du = 2 \, dx \implies dx = \frac{du}{2}
\][/tex]
4. Rewrite the integral in terms of [tex]\( u \)[/tex]:
Substitute [tex]\( u = 2x + 1 \)[/tex] and [tex]\( dx = \frac{du}{2} \)[/tex] into the integral:
[tex]\[
\int \frac{1}{2x + 1} \, dx = \int \frac{1}{u} \cdot \frac{du}{2}
\][/tex]
[tex]\[
= \frac{1}{2} \int \frac{1}{u} \, du
\][/tex]
5. Integrate with respect to [tex]\( u \)[/tex]:
The integral of [tex]\(\frac{1}{u}\)[/tex] with respect to [tex]\(u\)[/tex] is [tex]\(\ln|u|\)[/tex] (the natural logarithm of the absolute value of [tex]\( u \)[/tex]):
[tex]\[
\frac{1}{2} \int \frac{1}{u} \, du = \frac{1}{2} \ln|u| + C
\][/tex]
where [tex]\( C \)[/tex] is the constant of integration.
6. Substitute back [tex]\( u = 2x + 1 \)[/tex]:
Replace [tex]\( u \)[/tex] with [tex]\( 2x + 1 \)[/tex]:
[tex]\[
\frac{1}{2} \ln|u| + C = \frac{1}{2} \ln|2x + 1| + C
\][/tex]
7. Simplify the expression:
Since [tex]\( 2x + 1 \)[/tex] is always positive for real values of [tex]\( x \)[/tex] (assuming the domain where it doesn't make the argument negative inside the logarithm), we can drop the absolute value:
[tex]\[
\frac{1}{2} \ln(2x + 1) + C
\][/tex]
So, the solution to the integral [tex]\(\int \frac{1}{2x + 1} \, dx\)[/tex] is:
[tex]\[
\boxed{\frac{1}{2} \ln(2x + 1) + C}
\][/tex]
where [tex]\( C \)[/tex] is the constant of integration.
