So we all know about the common integral:

$$\int{\frac{dx}{\sqrt{x^2+1}}} = \ln{|x + \sqrt{x^2+1}|} + C$$

Out of boredom I've decided to try and solve the integral to see if I could get the RHS expression. I've started with $$x = \tan{u} $$ and that substitution lead me to the following solution: $$\int{\frac{dx}{\sqrt{x^2+1}}} = \frac{1}{2}\ln{|\frac{1+\sin{(\arctan(x))}}{1 - \sin{(\arctan(x))}}|} +C$$

I wasn't sure whether it's correct so I've typed it up in an online derivative calculator and indeed got the original integral. So I'm wondering how can the solution differ by this much? Or is there a method to convert my expression into the original? If there is such a method then I'm really not seeing it.


There is no difference. Note that $$ \sin(\theta)=\frac{\tan(\theta)}{\sqrt{1+\tan^2(\theta)}}. $$ Then, you have $\theta=\arctan(x)$ and you get $$ \ln\left|\frac{1+\frac{x}{\sqrt{1+x^2}}}{1-\frac{x}{\sqrt{1+x^2}}}\right| $$ and you are done.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.