1. ## Standard Integration

Q: $\int \frac{1}{x\sqrt{1-x^2}} \ \mathrm{d}x$

Is there something really simple that I am not seeing? How would this be done? Thanks in advance.

2. Originally Posted by Air
Q: $\int \frac{1}{x\sqrt{1-x^2}} \ \mathrm{d}x$

Is there something really simple that I am not seeing? How would this be done? Thanks in advance.
Apply a U-Substitution. $u=1-x^2$. You can take it from there.

3. Originally Posted by Chris L T521
Apply a U-Substitution. $u=1-x^2$. You can take it from there.
Hehe...I overlooked it too. You can't use $u$-substitution...give me a minute to think it over...

4. Originally Posted by Air
Q: $\int \frac{1}{x\sqrt{1-x^2}} \ \mathrm{d}x$

Is there something really simple that I am not seeing? How would this be done? Thanks in advance.
Apply a trig substitution $x=\sin\theta$.

Also, recall that $\int \csc\theta \,d\theta= -\ln|\csc\theta+\cot\theta|+C$

5. Hello,

You can make a substitution $t=\sqrt{1-x^2}$

$\implies dt=\frac{-x}{t} \ dx \implies dx=-\frac tx \ dt$

$\int \frac{1}{x \sqrt{1-x^2}} dx=\int \frac{-1}{x^2} dt=- \int \frac{1}{1-t^2} dt=-\frac 12 \int \frac{1}{1-t}+\frac{1}{1+t} dt$

(modulo my mistakes )

6. Make the substitution $z^2=1-x^2.$

(Moo beat me for just one minute.)