can someone please help?
I need to evaluate this indefinite integral:
$$\int \frac{x}{\sqrt{x^2+2}}dx$$ I tried using substitution for letting u = x, but I can't get past finding the antiderivative after that.
Thank you!
2
$\begingroup$
$\endgroup$
-
1$\begingroup$ This one $\int \frac{x}{\sqrt{x^2+2}}dx$? $\endgroup$ – 3SAT Dec 8 '15 at 21:25
-
$\begingroup$ Yes! Thank you! I couldn't format it the right way $\endgroup$ – Shavana.S Dec 8 '15 at 21:26
-
1$\begingroup$ try $u=\sqrt{x^2+2}$ $\endgroup$ – WW1 Dec 8 '15 at 21:28
-
$\begingroup$ $u=x$ is not very useful, try one that simplifies the argument of the square root, e.g., $u=x^2+2$. $\endgroup$ – Oskar Limka Dec 8 '15 at 21:29
-
$\begingroup$ I'd suggest writing the root as $\sqrt{2}\sqrt{\frac{x^2}{2}+1}$ then putting $\frac{x}{\sqrt{2}}=\sinh t$. $\endgroup$ – MickG Dec 8 '15 at 21:58
1
$\begingroup$
$\endgroup$
For this you want to set $u = x^2+2$ which means $du=2x*dx$. We see an $x$ in the integrand which is good but we don't see a $2$. We can make up for this by saying.
$$\int{\frac{x}{\sqrt{x^2+2}}} = \frac{1}{2}\int{\frac{2x}{\sqrt{x^2+2}}}$$
And now we can use our $u$ substitution to get.
$$\int{\frac{x}{\sqrt{x^2+2}}}=\frac{1}{2}\int{\frac{1}{\sqrt{u}}du}=\frac{1}{2}\frac{\sqrt{u}}{\frac{1}{2}}+C=\sqrt{u}+C$$
Substituting $x^2+2$ back in for $u$ we find that.
$$\int{\frac{x}{\sqrt{x^2+2}}}=\sqrt{x^2+2}+C$$
answered Dec 8 '15 at 23:14
1
$\begingroup$
$\endgroup$
Letting $u=x$ doesn't accomplish anything. Try letting $u=x^2+2$.
answered Dec 8 '15 at 21:28
0
$\begingroup$
$\endgroup$
write the integral inthe form $$\frac{1}{2}\int\frac{2x}{\sqrt{x^2+2}}dx$$ and set $$x^2+2=t$$
answered Dec 8 '15 at 21:28
