Question

integral ln x^1/3 dx

Answer 1

can someone help me

Answer 2

since this is an ln function, I would suggest trying integration by parts.

Answer 3

still there?

Answer 4

If you let u = x^(1/3), then u^3 = x, and 3u^2 du = dx. Then, \[\int\limits_{}^{} \ln x ^{1/3}dx = 3\int\limits_{}^{} u ^{2} \ln u du\] and you can do integration by parts.

Answer 5

For variables in your integration by parts: \[\int\limits_{}^{} w dz = wz - \int\limits_{}^{} zdw\] Let w = ln u and dz = 3u^2 du, Then, dw = u^(-1) du, and z = u^3. So, integrating by parts, you will have: \[u ^{3}\ln u - \int\limits_{}^{}u ^{3}(1/u)du\] This last integral is easy enough to integrate, so I leave that to you.

Answer 6

Once you integrate that last term, put the equation back in terms of x and then you're done.

Answer 7

No one is around, so I'll just finish the job for my own fun. The last expression for the whole integral expression comes down to:\[u ^{3}\ln u - \int\limits_{}^{}u ^{2}du\] which equals:\[u ^{3}\ln u - (1/3)u ^{3}\] which equals:\[x(\ln x ^{1/3} - 1/3)\] I went ahead and took the derivative and it goes back to ln x^(1/3) so it checks out and is verified.