Find int (x^3+4) / x^2 dx (please check my work)

[confused_07]

Find int (x^3+4) / x^2 dx

What did I do wrong? Answer should be (x^3-8) / (2x)

int (x^3+4) / x^2 dx
int x^3 dx + int 4 dx * int x^(-2) dx
(x^4/4) + 4x * [x^(-1) / -1] + C
[(1/4)x^4 + 4x + C] / x

Thanks....

 

[pka]

NO, no and again double no!
\(\displaystyle \L\int {\frac{{x^3 + 4}}{{x^2 }}dx = \int {\left( {x + 4x^{ - 2} } \right)dx} }\)

 

[skeeter]

What did I do wrong?

the algebra ...

\(\displaystyle \L \frac{x^3 + 4}{x^2} = \frac{x^3}{x^2} + \frac{4}{x^2} = x + 4x^{-2}\)

integrate the last expression ...

\(\displaystyle \L \frac{x^2}{2} - \frac{4}{x} + C = \frac{x^3}{2x} - \frac{8}{2x} + C = \frac{x^3 - 8}{2x} + C\)

 

[confused_07]

Why is it you are so wrapped up in trying to learn the new stuff that you miss the simple things while doing so? Sorry for the that.....