How do you evaluate $int (1/(3-5x)^2)dx$ for [1, 2]?

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Answer 1 · 2015-10-27T02:19:48

Use substitution with $u = 3-5x$

$int_1^2 1/(3-5x)^2dx$

Let $u = 3-5x$ , so that $du = -5 dx$ and
when $x=1$ , we get $u = -2$
when $x=2$ , we get $u = -7$

The integral becomes

$-1/5 int_-2^-7 1/u^2 du = -1/5 [-1/u]_-2^-7$

$= -1/5[-1/(-7) - ( - 1/(-2)) = -1/5[1/7 - 1/2]$

$= -1/5[-5/14] = 1/14.$

How do you evaluate int (1/(3-5x)^2)dxint (1/(3-5x)^2)dx for [1, 2]?