I need some help with an Integral (Substitution).

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 our expert's

answer on brainly

SEE EXPERT ANSWER

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

A community for students.

I need some help with an Integral (Substitution).

Mathematics
I got my questions answered at brainly.com in under 10 minutes. Go to brainly.com now for free help!
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

\[\int\limits_{?}^{?}x^2*\sqrt{3x^2+4}dx\] 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.
its 3x^2+4 in sqrt but u put u = 2x^3+4.....
Sry I got the first line wrong. This is the integral I am supposed so solve. \[\int\limits_{?}^{?}x^2*\sqrt{2x^3+4} dx\]

Not the answer you are looking for?

Search for more explanations.

Ask your own question

Other answers:

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

Not the answer you are looking for?

Search for more explanations.

Ask your own question