∫ (5 / x² -4/x) dx
I'm confused as to what to do. Do I begin by taking the anti derivative and if so how it is done?

Question

∫ (5 / x² -4/x) dx
I'm confused as to what to do. Do I begin by taking the anti derivative and if so how it is done?

anonymous · Answer

5x^-2 - 4x^-1
took the anti derivative and got 5x^-1/-1 - 4x^0/0. Do I ignore the second term since it is undefined?

kropot72 · Answer

The second term in your putative answer is $$\infty$$. Then you should add the constant of integration as the third term of the answer.

anonymous · Answer

Im confused as to how an integer divided by zero is equivilant to infinity?

kropot72 · Answer

0/0 is indeterminate (undefined). However any real number divided by zero will give the result of infinity - subject to a positive or negative sign.

kropot72 · Answer

Sorry. The second term in your putative answer should be:
$$-(4\times \ln x)$$

anonymous · Answer

ohh okay that makes much more sense

callisto · Answer

∫ (5 / x² -4/x) dx
= ∫ (5x^-2 -4/x) dx
=∫ (5x^-2 )dx -∫ (4/x) dx
= 5 [x^(-2+1)] / (-2+1) -4ln|x| +C
= -5 /x - 4ln|x| +C

kropot72 · Answer

You're welcome. The integral of 1/x does not follow the usual rule for integration.

anonymous · Answer

Thanks guys I really appreciate youre help!