How do you find the integral
#intsqrt(x^2-1)/xdx# ?

Question

How do you find the integral
 #intsqrt(x^2-1)/xdx# ?

Emma Aldridge · Accepted Answer

The answer is #=sqrt(x^2-1)-arctan(sqrt(x^2-1))+C#

We need

#int(dx)/(x^2+1)=arctanx+C#

Perform this integral by substitution

Let #u=sqrt(x^2-1)#, #=>#, #x^2=u^2+1#

#du=(2xdx)/(2sqrt(x^2-1))=(xdx)/(sqrt(x^2-1))#

Therefore,

#int(sqrt(x^2-1)dx)/(x)=intsqrt(x^2-1)/x*sqrt(x^2-1)/x*dx#

#=int((x^2-1)du)/(x^2)#

#=int(u^2)/(u^2+1)*du#

#=int(u^2+1-1)/(u^2+1)du#

#=intdu-int(du)/(u^2+1)#

#=u+arctan(u)#

#=sqrt(x^2-1)-arctan(sqrt(x^2-1))+C#

Scarlett Allison · Answer

undefined To find the integral $ \int \frac{\sqrt{x^2 - 1}}{x} \, dx $, you can use trigonometric substitution. Let $ x = \sec(\theta) $, then $ dx = \sec(\theta) \tan(\theta) \, d\theta $. Substituting these into the integral gives:

$$ \int \frac{\sqrt{x^2 - 1}}{x} \, dx = \int \frac{\sqrt{\sec^2(\theta) - 1}}{\sec(\theta)} \sec(\theta) \tan(\theta) \, d\theta $$

Simplify the expression under the square root:

$$ \sqrt{\sec^2(\theta) - 1} = \sqrt{\tan^2(\theta)} = \tan(\theta) $$

So the integral becomes:

$$ \int \tan^2(\theta) \, d\theta $$

Now, you can integrate $ \tan^2(\theta) $ using trigonometric identities. After integration, substitute back $ x = \sec(\theta) $ to get the final result.

How do you find the integral #intsqrt(x^2-1)/xdx# ?