Calculate the following integral:
$$\int \frac{1}{(1-x^2)\sqrt{1+x^2}} dx$$
I believe I need to choose a good substitution, but the problem is that I haven't found it yet.
Thank you!
Asked
Active
Viewed 55 times
2 Answers
1
HINT:
To evaluate integrals of the form $\displaystyle\int\frac{1}{(lx^2+m)\sqrt{ax^2+b}}\,dx,$ substitute $\sqrt{ax^2+b}=xz$ for another function $z$. This would lead to the primitive yielding an $\arctan$ function.
StubbornAtom
- 17,052
1
HINT:
Set $x=\tan y\implies dx=?$
$$\int \frac1{(1-x^2)\sqrt{1+x^2}} dx=\int\dfrac{\sec y\ dy}{1-\tan^2y}=\int\dfrac{\cos y\ dy}{1-2\sin^2y}$$
Set $\sin y=u$
Or directly, $u=\dfrac x{\sqrt{1+x^2}}$
lab bhattacharjee
- 274,582
-
Can you, please, explain how you got what's on the last line? – George R. Nov 27 '16 at 18:14
-
@GeorgeR., $$\sin y=\dfrac{\tan y}{\sec y}=\dfrac{\tan y}{\sqrt{1+\tan^2y}}$$ for $-\dfrac\pi2\le y\le\dfrac\pi2$ – lab bhattacharjee Nov 27 '16 at 18:19