Following the hint, we start off by setting
Then we have that
and more importantly that
Thus, substituting these values in yield
Via the third hint, we can simplify this last integral to
(one could find this expression by explicitly doing the long division though on an exam you probably want to try to save some time). The first integral is
(we'll add the constant in later) and the second integral we evaluate by partial fractions.
This gives the expression
Plugging in yields
and so . Plugging in yields
and so . Hence
(we'll add the constant in later). Hence, applying everything, we have
completing the question.
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