Question

Calculus 2
If we have
∫ (sec² B / tan² B) sec B tan B dB
And u = secB
Du = secBtanBdB
And we get ∫(u^2)/(u^2-1) du
How do we go from there to :
∫ ( 1 + (1/(u^2-1) du
Can somebody explain to me fully why it became that last expression?

Answer 1

A simple way to see what was done is to add and subtract -1 from the numerator:

$\dfrac{u^2-1+1}{u^2-1}=\dfrac{u^2-1}{u^2-1}+\dfrac1{u^2-1}=1+\dfrac1{u^2-1}$

(provided that $u^2-1\neq0$, or $u\neq\pm1$)

###

Suppose you had a slightly more complex integrand, like

$\dfrac{u^3}{u^2-1}=u+\dfrac u{u^2-1}$

How do we know that? (Assume we don't already know the previous result, so that it's not just a matter of multiplying both sides by $u$.) Simple polynomial division:

$u^3=\boxed{u}_{\,q}\cdot u^2$, and $u(u^2-1)=u^3-u$. Subtracting this from $u^3$ gives a remainder of $u^3-(u^3-u)=\boxed{u}_{\,r}$, so

$\dfrac{u^3}{u^2-1}=\boxed{u}_{\,q}+\dfrac{\boxed{u}_{\,r}}{u^2-1}$

(where $q$ and $r$ denote "quotient" and "remainder")