How do you find the antiderivative of #int 1/(3x-7)^2 dx# from [3,4]?

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Answer 1 · 2016-12-06T17:14:43

We can rewrite this as

#int(3x - 7)^-2dx#

We let #u = 3x - 7#, then #du = 3(dx)#, and #dx = (du)/3#.

#=>int_3^4(u)^-2(1/3)du#

#=>1/3int_3^4(u)^-2du#

#=>1/3(-1/2u^-1)|_3^4#

#=>-1/(6u)|_3^4#

#=>-1/(6(3x- 7))|_3^4#

#=>-1/(18x - 42)|_3^4#

We now evaluate using #int_a^b(f(x)) = F(b) - F(a)#, where #F(x)# is the antiderivative of #f(x)#.

#=>-1/(18(4) - 42) - (-1/(18(3) - 42))#

#=>-1/30 + 1/12#

#=>1/20#

Hopefully this helps!