Question

integrate 36(4x^2-9)^-1

Answer 1

\[\int\limits_{}^{}\frac{36}{4x^{2}-9}dx\]
is it?

Answer 2

Just use the fact that
\[\frac{36}{4x^{2}-9}=\frac{6}{2x-3}-\frac{6}{2x+3}\]

Answer 3

yes , then how to integrate it ? one by one or what?

Answer 4

Well, the integral sign distributes over addition (and over subtraction)
And use the fact that for any nonzero a, b, and c
\[\int\limits\limits_{}^{}\frac{c}{ax+b}dx=\frac{c}{a}\ln|ax+b|+K, \ \ \ K \in \mathbb{R}\]

Answer 5

what formula is that ? never seen it .

Answer 6

\[\int\limits_{}^{}\frac{c}{ax+b}dx\]Let u=ax+b
then du = adx
\[\int\limits\limits_{}^{}\frac{c}{ax+b}dx=c\int\limits\limits_{}^{}\frac{1}{ax+b}dx=\frac{c}{a}\int\limits\limits_{}^{}\frac{1}{u}du\]
\[=\frac{c}{a}\ln \left| u \right|+K\]
\[=\frac{c}{a}\ln \left| ax+b \right|+K\]

Answer 7

okay ! thanks :)