hi how do i integrate the following (what is the general format etc): 1/((3x+1)^2)
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Originally Posted by taurus hi how do i integrate the following (what is the general format etc): 1/((3x+1)^2) $\displaystyle \int \frac{dx}{(3x + 1)^2}$ Let $\displaystyle u = 3x + 1 \implies du = 3dx$, so $\displaystyle \int \frac{dx}{(3x + 1)^2} = \frac{1}{3} \int \frac{du}{u^2}$ You can do it from there. -Dan
does this seem right" => 1/2 (U^-1/-1) => -1/(2U) ?
Originally Posted by taurus does this seem right" => 1/2 (U^-1/-1) => -1/(2U) ? it's 1/3 not 1/2
ooop yea thanks
Originally Posted by taurus ooop yea thanks and of course, you should remember to replace u with 3x + 1 when done
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