Question

integral of xsqrt(x^2 + 1)

Answer 1

\[\large \int x\sqrt{x^2+1}dx\]Could you try to apply some substituion here? :D

Answer 2

I'm really bad at u substitutions. I always get confused as to which is substituted. Is it u=x^2+1 and du=2x? then I replace the x^2+1 with u and x with du? But then do I need to put a 1/2 out infront of the integral?

Answer 3

use
\[u=x^2+1\]
\[du=2x\]
You will need to multiply and divide by 2 and you're set.

Answer 4

is the final answer 1/3(x^2+1)^(3/2) +c?????? :)

Answer 5

Basically, the substitution rule is to take a complicated thing and use a letter to define that, and then, you could use derivative to get the dx out and substitution wrt the letter you assigned to that thing

Answer 6

Yes it is Heather, that's right.

Answer 7

Yay! Thank you both so much!

Answer 8

\[\large \int x\sqrt{x^2+1}dx
\\\text{Let} ~u=x^2+1;~~du = 2x~dx
\\\frac{du}{2}=xdx

\\\large \int \frac{1}{2}\sqrt{u}du=\frac{1}{2}\int \sqrt{u}du=\frac{1x^{\frac{3}{2}}}{3}+c=\frac{(x^2+1)^{\frac{3}{2}}}{3}+c
\]yup, correct :)

Answer 9

Thank you!!