Tìm nguyên hàm của hàm số: ∫dx2x2x+1

Question

\(\int \frac{{\rm d}x}{2x\sqrt{2x+1} }\)

Answer 1

Đặt \(t=\sqrt{2x+1} \Rightarrow t^{2} =2x+1\Rightarrow 2x=t^{2} -1\Rightarrow {\rm d}x=t{\rm d}t\), do đó:
\(\int \frac{{\rm d}x}{2x\sqrt{2x+1} } =\int \frac{t{\rm d}t}{\left(t^{2} -1\right)t}\)

\(=\int \frac{1}{t^{2} -1} {\rm d}t=\frac{1}{2} \int \left(\frac{1}{t-1} -\frac{1}{t+1} \right){\rm d}t\)

\(=\frac{1}{2} \left(\ln \left|t-1\right|-\ln \left|t+1\right|\right)+C\)
\(=\frac{1}{2} \left(\ln \left|\sqrt{2x+1} -1\right|-\ln \left|\sqrt{2x+1} +1\right|\right)+C.\)