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Answer 1 · 2017-10-22T21:59:13

$\frac{- 1}{3 (3 x + 7)} + C$ $(-1)/(3(3x+7))+C$

$\int \frac{1}{{(3 x + 7)}^{2}} d x$ $int1/(3x+7)^2dx$

We can rewrite this:

$= \int {(3 x + 7)}^{- 2} d x$ $=int(3x+7)^-2dx$

Let's use the substitution $u = 3 x + 7$ $u=3x+7$ . This implies that $d u = 3 d x$ $du=3dx$ .

$= \frac{1}{3} \int {(3 x + 7)}^{- 2} (3 d x)$ $=1/3int(3x+7)^-2(3dx)$

$= \frac{1}{3} \int u^{- 2} d u$ $=1/3intu^-2du$

Now we can use the rule $\int u^{n} d u = \frac{u^{n + 1}}{n + 1} + C$ $intu^ndu=u^(n+1)/(n+1)+C$ , where $n \neq - 1$ $n!=-1$ .

$= \frac{1}{3} (\frac{u^{- 1}}{- 1}) + C$ $=1/3(u^-1/(-1))+C$

$= \frac{- 1}{3 u} + C$ $=(-1)/(3u)+C$

$= \frac{- 1}{3 (3 x + 7)} + C$ $=(-1)/(3(3x+7))+C$