Question

integrate x sqrt(2x-1)

Answer 1

i get 1/3(2x-1)^3/2 + C

Answer 2

is that correct?

Answer 3

That doesn't look right. Remember that there is another integral after performing integration by parts. I think you forgot to integrate the last part. That is, you got up to this step:
\[\int\limits u \ dv = uv - \int\limits v \ du\]
But the result you came to (above) does not take into account the final integral, that is, this part:
\[\int\limits v \ du\]

Answer 4

i'm suppose to solve using substitution

Answer 5

u=2x-1 ?

Answer 6

This is what I assume you've done so far (based on your answer above):
\[\int\limits x \sqrt{2x-1} \ dx\]
Using integration by parts:
\[u = x \\ du = dx \\ dv = \sqrt{2x-1} \\ v = \frac{1}{3} (2x-1)^{\frac{3}{2}}\]
Applying this to the integration by parts formula (which I posted above):
\[\int\limits x \sqrt{2x-1} \ dx = x \frac{1}{3} (2x-1)^{\frac{3}{2}} - \int\limits \frac{1}{3} (2x-1)^\frac{3}{2} \ dx\]
You said you performed substitution to get to here. I'm not sure how that would work, since, if you chose u = 2x-1, then du = 2 and you wouldn't be able to substitute for x (unless you performed some complex substitution).
Now carry out the integration in the last equation, this time using substitution, with, as you mentioned above, u = 2x-1 (this time it should work).

Answer 7

it looks like i have to use IBP. you're right I was running into the problem of not having anything for x. my instructions just say use substitution as if IBP doesn't exist yet.

Answer 8

anyways, thanks for the help!

Answer 9

The work and explanation seems correct and according to my resource i think the answers are correct as well