[tex]F(x)=\int x^3\cdot lnx\ dx= \int\left(\dfrac{x^4}4\right)'\cdot lnx\ dx=\dfrac{x^4}4\cdot lnx-\int\dfrac{x^4}4\cdot\dfrac{1}x\ dx=\\\\=\dfrac{x^4}4\cdot lnx-\dfrac{x^4}{16}+C.\ Pe\ C\ il\ afl\breve{a}m\ din\ condi\c{t}ia\ F(1)=0:\\\\F(1)=-\dfrac1{16},\ deci\ C=-\dfrac{1}{16}.\\\\F(x)=\dfrac{x^4}4\cdot lnx-\dfrac{x^4}{16}-\dfrac1{16}.[/tex]
Pentru aflarea primitivei am aplicat metoda integrării prin părți.
Green eyes.
