- #1
Whatupdoc
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i will use "\int" as a integral sign since latex is down.
\int (7)/(x^2-1)*dx
using partial fractions...
took out the 7...
7\int (1)/(x+1)(x-1)
A(x-1) + B(x+1) = 7
if x = 1, B=7/2
if x = -1, A= -7/2
ok it's time to set up my integral function:
7\int -7/2(x-1) + 7\int 7/2(x-1)
take out the 7...
14\int -1/2x-1 + 14\int 1/2x-2
now to solve the integrals
14(-ln(2x-2) + ln(2x-2)) + c
where did i go wrong?
\int (7)/(x^2-1)*dx
using partial fractions...
took out the 7...
7\int (1)/(x+1)(x-1)
A(x-1) + B(x+1) = 7
if x = 1, B=7/2
if x = -1, A= -7/2
ok it's time to set up my integral function:
7\int -7/2(x-1) + 7\int 7/2(x-1)
take out the 7...
14\int -1/2x-1 + 14\int 1/2x-2
now to solve the integrals
14(-ln(2x-2) + ln(2x-2)) + c
where did i go wrong?