anonymous
  • anonymous
I need some help with an Integral (Substitution).
Mathematics
katieb
  • katieb
See more answers at brainly.com
At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga. Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus. Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.

Get this expert

answer on brainly

SEE EXPERT ANSWER

Get your free account and access expert answers to this
and thousands of other questions

anonymous
  • anonymous
\[\int\limits_{?}^{?}x^2*\sqrt{3x^2+4}dx\] I started with \[u = 2x^3 +4\] \[du = 6x^2dx \rightarrow x^2dx=1/6 * du\] \[\int\limits_{?}^{?}\sqrt{u}*du/6 = 1/9 * \sqrt{u^3} = 1/9 * \sqrt{(2x^3+4)^3}\] But Wolfram got something else entirely but I fail to see my mistake.
hartnn
  • hartnn
its 3x^2+4 in sqrt but u put u = 2x^3+4.....
anonymous
  • anonymous
Sry I got the first line wrong. This is the integral I am supposed so solve. \[\int\limits_{?}^{?}x^2*\sqrt{2x^3+4} dx\]

Looking for something else?

Not the answer you are looking for? Search for more explanations.

More answers

hartnn
  • hartnn
i don't see any fault in your work.
anonymous
  • anonymous
Ups sry. There is no mistake. Wolfram just pulls a 2 out of the squareroot term.
anonymous
  • anonymous
Thanks for the douple checking.^^ I am sorry.
hartnn
  • hartnn
no problem welcome :)

Looking for something else?

Not the answer you are looking for? Search for more explanations.