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!

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