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using integration table evaluate the following integral sqrt(9x^2-1)
I just need to know how to start this off, i tried u substitution:
u=9x^2-1 du=18xdx integral:u^(1/2)du/18x. but I don't know how to get rid of the x,
So then from there i tried to use from the integration table integral:udv = uv-(integral vdu)
u=u^1/2 dv=1/18x du=u^(3/2)/(3/2) v=
I didn't know how to go about that
I just need to know how to start this off, i tried u substitution:
u=9x^2-1 du=18xdx integral:u^(1/2)du/18x. but I don't know how to get rid of the x,
So then from there i tried to use from the integration table integral:udv = uv-(integral vdu)
u=u^1/2 dv=1/18x du=u^(3/2)/(3/2) v=
I didn't know how to go about that