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take out 7 as common factor from numerator and put denominator =x and diff

\[\int\limits_{}\frac{7x^4 +112x^3+63}{x^5+20x^4+45(??)+2}dx\]

I have gotten to
7(x^4 + 16x^3 + 9)/(x^4 +16x^3 + 9)

the 45 i left out the x
should be 45x

when diff. denominator we get, 5x^4+80x^3 +45 or 5(x^4+16x^3+9)

that should be the du isn't it???

(5x^4+80x^3 +45)dx= dt
5(x^4+16x^3+9)dx=dt
(x^4+16x^3+9)dx=dt/5

mutilpy the top and bottom by 5/7 to get the top as the derivative of the bottom

\[\int\limits_{} \frac{\frac{5}{7}7(x^4+16x^3+9)}{\frac{5}{7}(x^5 +20x^4 +45x +2)}dx\]

didn't we need to substitue somewhere??

I did, really:
u = x^5+20x^4+45x+2
du = 5(x^4 +16^3 +9)

We turned it into the form:\[\frac{7}{5}\int\limits_{}\frac{du}{u}\]

between the 9 and (x^4 they are not on the same line th (x^4 is next to the fraction

\[7 \int\limits_{}\frac{\frac{1}{x^5}+16x^3+9}{x^4+4x^3+2}dx\] like this?

frac{top}{bottom} is how you get the line between them like that

I guess it is called differentiate
1/x^5+16x^3+9 then next to this fraction is (x^4+4x^3+2)dx

1frac{top}{bottom}x^5 +16x^3 +9 x^4+4x^3+2dx

never mind that didn't work

\[\int\limits_{}\frac{(x^4+4x^3+2)}{x^5+16x^3+9}dx\] is what the amounts to then...

\[u = x^5 +16x^3 +9; du=5x^4+48x^2\]

\[7\int\limits_{?}^{?}\] then the equation you have