Đặt \(\left\{\begin{array}{l} {u=\ln ^{2} x} \\ {{\rm d}v=x^{3} {\rm d}x} \end{array}\right. \Rightarrow \left\{\begin{array}{l} {{\rm d}u=2\ln x\frac{{\rm d}x}{x} \, } \\ {v=\frac{x^{4} }{4} } \end{array}\right. .\)
Khi đó:\(\int x^{3} \ln ^{2} x\, {\rm d}x =\frac{x^{4} }{4} \ln ^{2} x-\frac{1}{2} \int x^{3} \ln x\, {\rm d}x=\frac{x^{4} }{4} \ln ^{2} x-\frac{1}{2} J.\)
Tính \(J=\int x^{3} \ln x\, {\rm d}x\)
Đặt \(\left\{\begin{array}{l} {a=\ln x} \\ {{\rm d}v=x^{3} {\rm d}x} \end{array}\right. \Rightarrow \left\{\begin{array}{l} {{\rm d}a=\frac{{\rm d}x}{x} } \\ {v=\frac{x^{4} }{4} } \end{array}\right. .\)
Khi đó:\(J=\frac{x^{4} }{4} \ln x-\frac{1}{4} \int x^{3} {\rm d}x=\frac{x^{4} }{4} \ln x-\frac{x^{4} }{16} +C.\)
Vậy: \(\int x^{3} \ln ^{2} x\, {\rm d}x =\frac{x^{4} }{4} \ln ^{2} x-\frac{x^{4} }{8} \ln x+\frac{x^{4} }{32} +C.\)