Integration by parts 
I have done the 1st part of question and the answer I got is $\frac{5e^4 - 1}{32} $which I verified from the calculator too. But I am confused how to approach to the deducing part using previous result(since it is stated HENCE ). Any help is greatly appreciated. 

$$\int _{ 1 }^{ e }{ { x }^{ 3 }{ \left( \ln { x }  \right)  }^{ 2 } } dx=\frac { 1 }{ 4 } \int _{ 1 }^{ e }{ { \left( \ln { x }  \right)  }^{ 2 } } d\left( { x }^{ 4 } \right) ={ \frac { { x }^{ 4 }{ \left( \ln { x }  \right)  }^{ 2 } }{ 4 }  }_{ 1 }^{ e }-\frac { 2 }{ 4 } \int _{ 1 }^{ e }{ { x }^{ 3 } } \ln { x } dx=\\ =\frac { { e }^{ 4 } }{ 4 } -\frac { 1 }{ 8 } \left( \int _{ 1 }^{ e }{ \ln { x }  } d\left( { x }^{ 4 } \right)  \right) =\frac { { e }^{ 4 } }{ 4 } -\frac { 1 }{ 8 } \left( { { x }^{ 4 }\ln { x }  }_{ 1 }^{ e }-\int _{ 1 }^{ e }{ { x }^{ 3 }dx }  \right) =\\ =\frac { { e }^{ 4 } }{ 4 } -\frac { 1 }{ 8 } \left( { e }^{ 4 }-{ \frac { { x }^{ 4 } }{ 4 }  }_{ 1 }^{ e } \right) =\frac { { e }^{ 4 } }{ 4 } -\frac { 1 }{ 8 } \left( { e }^{ 4 }-\frac { { e }^{ 4 } }{ 4 } +\frac { 1 }{ 4 }  \right) =\frac { 8{ e }^{ 4 }-4{ e }^{ 4 }+{ e }^{ 4 }-1 }{ 32 } =\frac { 5{ e }^{ 4 }-1 }{ 32 } $$
