Antiderivative of 1/dx

[scrum]

This seems really simple but I don't know what it is.

it occurs in a u substitution in

i'm saying that 1/du integrates to u but my answer is wrong and I think this is the step that is messing me up.

The answer I am getting is -4 { sqrt(t) * arctan([sqrt(t)]) - .5ln(t+1) } + C I use U sub, then integration by parts on the U subbed thing, and then U sub again on the second part of the integration by parts.

 

[o_O]

Your answer is almost there. There's just something wrong with the coefficients. I'm not sure how you obtained the -4 and the .5. You started off with a -8 and I don't think you halve it ... If possible, post what you have and we can figure out what may be happening.

 

[Dr. Flim-Flam]

On the above integral if you let x = t^(1/2), you will get

-16{integral(arctan(x)dx). Then use integration by parts and then sub t^(1/2) for x

and you'll get 8ln(1+t) - 16t^(1/2)arctan(t^(1/2)).+ C.