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- Identify integral: Identify the integral that needs to be solved: .
- Constant multiple rule: Use the constant multiple rule to take the constant out of the integral: .
- Integration by parts: Apply integration by parts, where and . Then we need to find and .
- Find : Differentiate to find : $du=\left(\frac{\(1\)}{\(2\)x+\(1\)}\right) \cdot \(2\)dx = \left(\frac{\(2\)}{\(2\)x+\(1\)}\right)dx.
- Find \(v\): Integrate \(dv\) to find \(v\): \(v=x\).
- Integration by parts formula: Plug \(u\), \(du\), and \(v\) into the integration by parts formula: \(\int u\,dv = uv - \int v\,du\).
- Substitute values: Substitute the values into the formula: \(4[\ln(2x+1)x - \int(x\left(\frac{2}{2x+1}\right)dx)]\).
- Simplify integral: Simplify the integral: \(4[(\ln(2x+1)x) - \int(\frac{2x}{2x+1})dx]\).
- Divide \(2x\): Divide \(2x\) by \((2x+1)\) in the integral: \(4[\ln(2x+1)x - \int(1 - \frac{1}{2x+1})dx]\).
- Split integral: Split the integral into two parts: \(4[\ln(2x+1)x - ( \int 1\,dx - \int \frac{1}{2x+1}\,dx)]\).
- Integrate both parts: Integrate both parts: \(4\left[\left(\ln(2x+1)x\right) - \left(x - \ln|2x+1| + C\right)\right].\)
- Combine like terms: Combine like terms and multiply by \(4\): \(4\ln(2x+1)x - 4x + 4\ln|2x+1| + C\).
- Final solution: Notice that \(4\ln(2x+1)x\) and \(4\ln|2x+1|\) are similar terms and can be combined: \(4x(\ln(2x+1) - 1) + C\).
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