Question

Show that,

Answer 1

\[\int\limits_{}^{}\frac{2-x}{4x^{2}+4x-3}=-\frac{1}{8}\ln \left| 4x^{2}+4x-3 \right|+\frac{5}{16}\ln \left| \frac{2x-1}{2x+3} \right|+C\]

Answer 2

at first glance i think i will approach it by factoring the denominator and then solve with partial fraction. but the answer doest seems to give me the RHS

Answer 3

did you try completing the square of the denominator?

Answer 4

o i try 1st

Answer 5

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

Answer 6

o.o so i separate them?

Answer 7

how did you get that 1/4?

Answer 8

oo omg ererer thr's a 4 infront of the bracket

Answer 9

yeah, i factored that 4 out of the integral so now i have 1/4*integral

Answer 10

this is what i have so far...

Answer 11

so its \[\frac{1}{2}\int\limits_{}^{}\frac{1}{u^{2}-1}-\frac{1}{4}\int\limits_{}^{}\frac{u-\frac{1}{2}}{u^{2}-1}\]?