Tính tích phân sau: I=∫x/(√x^2+1) +xdx

Question

Tính tích phân sau:

\(I=\int _{0}^{1}\frac{x}{\sqrt{x^{2} +1} +x} {\rm d}x .  \)

Answer 1

\(I=\int _{0}^{1}\frac{x}{\sqrt{x^{2} +1} +x} {\rm d}x  \)
Đặt  \( t=\sqrt{x^{2} +1} +x\Rightarrow t-x=\sqrt{x^{2} +1} \Rightarrow t^{2} -2tx=1\)

\(\Rightarrow x=\frac{t^{2} -1}{2t} \Rightarrow {\rm d}x=\frac{t^{2} +1}{2t^{2} } {\rm d}t\)

Đổi cận: \(x=0\Rightarrow t=1;\, \, x=1\Rightarrow t=1+\sqrt{2} \)
\(\int _{0}^{1}\frac{x}{\sqrt{x^{2} +1} +x} {\rm d}x =\int _{1}^{1+\sqrt{2} }\left(\frac{t^{2} -1}{2t} .\frac{\left(\frac{t^{2} +1}{2t^{2} } \right)}{t} \right){\rm d}t =\int _{1}^{1+\sqrt{2} }\left(\frac{t^{4} -1}{4t^{4} } \right){\rm d}t  \)
\(=\int _{1}^{1+\sqrt{2} }\left(\frac{1}{4} -\frac{1}{4t^{4} } \right){\rm d}t =\left. \left(\frac{1}{4} t+\frac{1}{12t^{3} } \right)\right|_{1}^{1+\sqrt{2} } =\frac{-1+2\sqrt{2} }{3} -\frac{1}{3} =\frac{-2+2\sqrt{2} }{3}  \)