-1
$\begingroup$

I don't really know where to start with this.

$$\int \sqrt{x^2+y^2+1}\quad dx$$

$\endgroup$

closed as off-topic by Mark Fantini, Davide Giraudo, symmetricuser, Chappers, Andrew D. Hwang Mar 31 '15 at 22:54

This question appears to be off-topic. The users who voted to close gave this specific reason:

  • "This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – Mark Fantini, Davide Giraudo, symmetricuser, Chappers, Andrew D. Hwang
If this question can be reworded to fit the rules in the help center, please edit the question.

  • $\begingroup$ What have you tried, or at least, what are your thoughts? Why does it look hard? Is it the $y$? $\endgroup$ – mickep Mar 31 '15 at 18:09
0
$\begingroup$

Take $x = \sqrt{y^2+1} \sinh u$. This yields

$$ \left (y^2+1 \right )\int \cosh^2 u du $$

which is more straight-forward to compute.

$\endgroup$
  • $\begingroup$ Is using hyperbolic trig integration easier?? Op might not even know how to evaluate that. $\endgroup$ – Zach466920 Mar 31 '15 at 17:32
  • $\begingroup$ @Zach466920 Yes it is. If worst comes to worst he can just rewrite it in terms of its exponential arguments and integrate the function that way. If he doesn't know hyperbolic trig identities, this wouldn't be such a bad time to learn them, they're pretty easy. Either way, it's not your responsibility to speak for the OP. Using your trig identity is quite a bit harder since you end up having to integrate $\sec^3\theta$. $\endgroup$ – Mnifldz Mar 31 '15 at 17:39
  • $\begingroup$ Considering the nature of the question, I think that was a fair assumption. Mabye you should add the definition of cosh(u) to the answer. $\sec^3 \theta$ can evaluated by integration by parts, $\cosh^2 u$ can be evaluated by appyling another trig identity and integration. You get the same answer, with similar amounts of work. $\endgroup$ – Zach466920 Mar 31 '15 at 17:47
  • $\begingroup$ I think maybe you want $x=\sqrt{y^2+1}\sinh u$. $\endgroup$ – user84413 Mar 31 '15 at 21:25
  • $\begingroup$ @user84413 Indeed I did. Thank you. $\endgroup$ – Mnifldz Mar 31 '15 at 21:40
0
$\begingroup$

Hint: Let $D=y^2+1$, because y is an independent variable, then use trignometric substitution.

$\endgroup$
  • $\begingroup$ Which trig identity? $\endgroup$ – picaposo Mar 31 '15 at 17:33
  • $\begingroup$ @picaposo You should use $x=tan(\theta)$. Take the derivative with respect to t. Cancel the resulting dt's and sub dx for $(stuff) \cdot d \theta$, where stuff is the correction factor from manipulating $x=tan(\theta)$ into a form with dx. $\endgroup$ – Zach466920 Mar 31 '15 at 17:35
-1
$\begingroup$

Well, you may easily solve this one without any substitutions - just apply inegration by parts. Take $~~u = \sqrt{1+y^2+x^2}~~$ and $~~ dv = dx ~~$. Then you'll get that

$I = \int \sqrt{1+y^2+x^2} ~~ dx = x \sqrt{1 + y^2 + x^2} - \int {\frac{x^2}{1+y^2+x^2}} ~~ dx = x \sqrt{1 + y^2 + x^2} - \int {(\sqrt{1+y^2+x^2} - \frac{1}{1+\frac{x^2}{1+y^2}})} ~~ dx = x \sqrt{1 + y^2 + x^2} - I + \int{\frac{1}{1+\frac{x^2}{1+y^2}})} $

The last one is a standard integral. All you need is to "extract" I.

$\endgroup$

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