Tích phân \(\int\limits^4_0\frac{4x-1}{\sqrt{2x+1}+1}\text{d}x\) bằng
\(\frac{10}{3}+\ln2\). \(\frac{22}{3}+\ln2\). \(\frac{-22}{3}+\ln2\). \(\frac{-10}{3}+\ln2\). Hướng dẫn giải:Đặt \(t=\sqrt{2x+1}+1\) ⇒ \(\left(t-1\right)^2=2x+1\)
=> \(2\left(t-1\right)\text{d}t=2\text{d}x\) và \(x=\frac{\left(t-1\right)^2-1}{2}\)
\(I=\int\limits^4_0\frac{4x-1}{\sqrt{2x+1}+1}\text{d}x=\int\limits^4_2\frac{4\frac{\left(t-1\right)^2-1}{2}}{t}\left(t-1\right)\text{d}t\)
\(=\int\limits^4_2\frac{\left(2t^2-4t-1\right)\left(t-1\right)}{t}\text{d}t\)
\(=\int\limits^4_2\frac{2t^3-6t^2+3t+1}{t}\text{d}t\)
\(=\int\limits^4_2\left(2t^2-6t+3+\frac{1}{t}\right)\text{d}t\)
\(=\left(\frac{2t^3}{3}-\frac{6t^2}{2}+3t+\ln t\right)|^4_2\)
\(=\frac{22}{3}+\ln2\).