These problems are destroying my mind.
\int 1/((10x-3)^2)
So far I've... \int u^2 1/udu, u=10x-3
After this is where my thoughts unravel.
Maybe...
(1/10) * (u^3)/3 = (1/10) * ((10x-3)^3)/3 ????
The next one is
\int (6x+1) * \sqrt{3x^2+x}dx where u is given and u = 3x^2+x
I realize the derivative of 3x^2+x is 6x+1 and this offers some kind of portent but I fail to see the next logical step.
Any help is appreciated.
I've worked out (with some help)
u =10x-3
du=10dx
1/10du=dx
1/10 * the integral of 1/u^2
I've taken this to mean
1/10 * -1/10x-3 or -1/100x-30
where -1/100x-30 is the final solution.
I'm having trouble deriving this, however. My answer comes out as 1/(100*x^2-60*x+9)