Question

table integral AGAIN

Answer 1

\[\int\limits \frac{ dx }{ x^2\sqrt{81x^2+25}
I'm thinking use # 28

Answer 2

\[\int\limits \frac{ dx }{ x^2\sqrt{81x^2+25} }\]

Answer 3

someone in my class told me the answer is \[\frac{ -\sqrt{81x^2+25} }{ 25x }\]
if that is true, can someone help me see how that worked out that way?

Answer 4

if a = 9x how does a^2u become 25x?

Answer 5

The form of #28 says
\[\int\frac{du}{u^2\sqrt{a^2+u^2}}=-\frac{\sqrt{a^2+u^2}}{a^2u}+C\]
Again, the problem with using tables for the sake of using tables poses a problem: you'll not always be given an integral that has the exact same form. In this case, you have \(\sqrt{a^2+\color{red}{\text{(something)}}u^2}\), while you need \(\sqrt{a^2+u^2}\).

To get around this, you have two decent options. The first is the derive an even more general form, and the other is to use a basic substitution. The second method is less tedious.

Suppose we substitute \(u=9x\), then \(u^2=81x^2\). Additionally, \(du=9\,dx\), or \(\dfrac{du}{9}=dx\). So, you have
\[\int \frac{ dx }{ x^2\sqrt{81x^2+25} }=\frac{1}{9}\int\frac{du}{\left(\dfrac{u}{9}\right)^2\sqrt{u^2+5^2}}=9\int\frac{du}{u^2\sqrt{u^2+5^2}}\]

Answer 6

i apologize, i got booted off trying to reply

Answer 7

ok, is this what it is...\[9\int\limits \frac{ du }{ u^2\sqrt{u^2+5} }= 9[\frac{ -\sqrt{25+81x^2} }{ 25(9x) }]\]=\[-\frac{ \sqrt{25+81x^2} }{ 25x }\]

Answer 8

or \[-\frac{ \sqrt{81x^2+25} }{ 25x }\]

Answer 9

?

Answer 10

Yep, it matches the answer you were given earlier.