To solve the integral [tex]\(\int \frac{(2 x+3)}{4 x^2+1} \, dx\)[/tex], let’s break it down into manageable parts and solve step-by-step.
1. Separate the Integrand:
First, separate the integral into two parts:
[tex]\[
\int \frac{2x}{4x^2 + 1} \, dx + \int \frac{3}{4x^2 + 1} \, dx
\][/tex]
2. Handle the First Integral:
Consider the integral [tex]\(\int \frac{2x}{4x^2 + 1} \, dx\)[/tex].
Notice that the derivative of the denominator [tex]\(4x^2 + 1\)[/tex] is [tex]\(8x\)[/tex]. Rewrite the numerator to match this, except for a constant factor:
[tex]\[
\int \frac{2x}{4x^2 + 1} \, dx = \frac{1}{4} \int \frac{8x}{4x^2 + 1} \, dx
\][/tex]
Now, recognize this as the natural logarithm form:
[tex]\[
\frac{1}{4} \int \frac{d(4x^2 + 1)}{4x^2 + 1}
\][/tex]
The integral of [tex]\(\frac{d(4x^2 + 1)}{4x^2 + 1}\)[/tex] is simply [tex]\(\ln|4x^2 + 1|\)[/tex]. Therefore:
[tex]\[
\frac{1}{4} \ln|4x^2 + 1|
\][/tex]
3. Handle the Second Integral:
Now, consider the integral [tex]\(\int \frac{3}{4x^2 + 1} \, dx\)[/tex].
To match the standard form of the arctangent integral, factor out the constant:
[tex]\[
3 \int \frac{1}{4x^2 + 1} \, dx = \frac{3}{2} \int \frac{1}{(2x)^2 + 1} \, dx
\][/tex]
Here, recognize this as the arctangent form:
[tex]\[
\frac{3}{2} \arctan(2x)
\][/tex]
4. Combine Both Results:
Combining the results from both parts, the full integral is:
[tex]\[
\int \frac{(2 x+3)}{4 x^2+1} \, dx = \frac{1}{4} \ln|4x^2 + 1| + \frac{3}{2} \arctan(2x) + C
\][/tex]
Simplifying the logarithmic term, it can be written as:
[tex]\[
\frac{1}{4} \ln\left(x^2 + \frac{1}{4}\right)
\][/tex]
Note that [tex]\(\ln|4x^2 + 1|\)[/tex] can also be expressed proportionally with [tex]\(\ln|x^2 + \frac{1}{4}|\)[/tex], making the structure clearer.
5. Final Answer:
Hence, the solution to the integral is:
[tex]\[
\int \frac{(2x+3)}{4x^2 + 1} \, dx = \frac{1}{4} \ln\left(x^2 + \frac{1}{4}\right) + \frac{3}{2} \arctan(2x) + C
\][/tex]
