indefinite integral of [(7x+3x^5+x^3)/x^3]dx

Question

anonymous · Answer

Divide each term of the numerator by x^3.

anonymous · Answer

$$\int\limits \frac{(7x+3x^5+x^3)}{x^3}dx$$
?

anonymous · Answer

divide by x^3 and convert to perfect sq

anonymous · Answer

yes that is the right equation and can you explain how i am able to divide by x^3?  I don't understand how i can do that

anonymous · Answer

$$\Huge \int\limits\limits \frac{(\frac{7x}{x^3}+\frac{3x^5}{x^3}+\frac{x^3}{x^3})}{1}dx$$

anonymous · Answer

so the 7x/x^3 would then be 7x^-2 ?

anonymous · Answer

$$\Huge \int\limits\limits\limits \frac{(\frac{7}{x^2}+\frac{3x^2}{1}+1)}{1}dx$$
now try converting to perfect square

anonymous · Answer

yep

anonymous · Answer

thank you! :)

anonymous · Answer

how do i convert to a perfect square?

anonymous · Answer

or u can let x^2=t

anonymous · Answer

If I may interject...that fraction can be simplified to the integral of  7/x^2  + 3x^2 + 1...and just integrate quickly (with the exception of the first one...but write first one as 7x^(-2). The other two is straightforward...x^3 and x, respectively. Why bother to change to a perfect square?