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

Question

Tính tích phân sau:

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

Answer 1

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

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

 + Tính \(\int _{0}^{1}\frac{x^{2} }{x^{3} +1} {\rm d}x . Đặt x^{3} +1=t\Rightarrow x^{2} {\rm d}x=\frac{1}{3} {\rm d}t\)

 Ta có \(\int _{0}^{1}\frac{x^{2} }{x^{3} +1} {\rm d}x =\frac{1}{3} \int _{1}^{2}\frac{1}{t} dt=\frac{1}{3} \left. \ln \left|t\right|\right| _{1}^{2} =\frac{1}{3} \ln 2\)

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