Tính tích phân sau: I = ∫( √ x^2 + 1 − √ 3 x + 1 ) x dx

Question

Tính tích phân sau:

\(I =\int _{0}^{1}\left(\sqrt{x^{2} +1} -\sqrt{3x+1} \right)x{\rm d}x .\)

 

Answer 1

\(I =\int _{0}^{1}\left(\sqrt{x^{2} +1} -\sqrt{3x+1} \right)x{\rm d}x\)

\(=\int _{0}^{1}x\sqrt{x^{2} +1} dx -\int _{0}^{1}x\sqrt{3x+1} {\rm d}x  \)
+ Tính \(\int _{0}^{1}x\sqrt{x^{2} +1} {\rm d}x\) .

Đặt \(\sqrt{x^{2} +1} =t\)

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

\(=\int _{1}^{\sqrt{2} }t^{2} {\rm d}t=\left. \frac{t^{3} }{3} \right|_{1}^{\sqrt{2} }  =\frac{-1+2\sqrt{2} }{3} . \)
+ Tính \(\int _{0}^{1}x\sqrt{3x+1} {\rm d}x \).

Đặt \(\sqrt{3x+1} =t\)

\(\Rightarrow 3x+1=t^{2} \Rightarrow dx=\frac{2}{3} t{\rm d}t\) và \(x=\frac{t^{2} -1}{3} \).
\(\int _{0}^{1}x\sqrt{3x+1} {\rm d}x \)

\(=\int _{1}^{2}\frac{t^{2} -1}{3} t.\frac{2}{3} t{\rm d}t=\frac{2}{9} \int _{1}^{2}\left(t^{4} -t^{2} \right){\rm d}t  =\left. \frac{2}{9} \left(\frac{t^{5} }{5} -\frac{t^{3} }{3} \right)\right|_{1}^{2} =\frac{116}{135} . \)
Vậy \(I =\int _{0}^{1}\left(\sqrt{x^{2} +1} -\sqrt{3x+1} \right)x{\rm d}x\)

\(=\int _{0}^{1}x\sqrt{x^{2} +1} {\rm d}x -\int _{0}^{1}x\sqrt{3x+1} {\rm d}x\)

\(=\frac{-1+2\sqrt{2} }{3} -\frac{116}{135} .\)