Question

integrate (4x^2-x+4)/(x^2+1)dx

Answer 1

\[\frac{4x^2-x+4}{x^2+1}=4-\frac{x}{x^2+1}\]

Answer 2

?

Answer 3

is that the fraction simplified?

Answer 4

yes...now use u substitution

Answer 5

im not completely sure how you simplified it like that

Answer 6

ah ok common denominators right

Answer 7

\[\frac{4x^2-x+4}{x^2+1}=\frac{4x^2+4-x}{x^2+1}\]

\[=\frac{4x^2+4}{x^2+1}-\frac{x}{x^2+1}\]
\[=\frac{4(x^2+1)}{x^2+1}-\frac{x}{x^2+1}=4-\frac{x}{x^2+1}\]

Answer 8

ok thx