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

Question

Tính tích phân sau:

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

Answer 1

\(I=\int _{-1}^{0}\left(\sqrt{x+1} -\sqrt{x^{2} +1} \right)xdx . \)
\(I=\int _{-1}^{0}\left(\sqrt{x+1} -\sqrt{x^{2} +1} \right)xdx =\int _{-1}^{0}\sqrt{x+1} xdx -\int _{-1}^{0}\sqrt{x^{2} +1} xdx \) 
Ta có:
\(I=\int _{-1}^{0}\sqrt{x+1} xdx \)

\( Đặt t=\sqrt{x+1} \Rightarrow dx=2tdt\)

 Đổi cận:\( \begin{array}{l} {x=-1\Rightarrow t=0} \\ {x=0\Rightarrow t=\sqrt{2} } \end{array}\)
\(I=2\int _{0}^{1}\left(t^{2} -1\right)t^{2} dtdx=\left(\frac{2}{5} t^{5} -\frac{2}{3} t^{3} \right) \left|\begin{array}{l} {1} \\ {0} \end{array}\right. =\frac{-4}{15}  \)

\(J=\int _{-1}^{0}\sqrt{x^{2} +1} xdx =\frac{1}{2} \int _{-1}^{0}\sqrt{x^{2} +1} \left(x^{2} +1\right)^{{'} } dx \)

\(=\frac{1}{3} \left(x^{2} +1\right)^{\frac{3}{2} } \left|\begin{array}{l} {0} \\ {-1} \end{array}\right. =\frac{1}{3} -\frac{2\sqrt{2} }{3} \)

 Vậy: \(\int _{-1}^{0}\left(\sqrt{x+1} -\sqrt{x^{2} +1} \right)xdx \)

\(=-\frac{4}{15} -\left(\frac{1}{3} -\frac{2\sqrt{2} }{3} \right)=\frac{2\sqrt{2} }{3} -\frac{3}{5} \)