How do you integrate ${(5 x - 3)}^{2} d x$ $(5x-3)^2dx$ ?

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How do you integrate ${(5 x - 3)}^{2} d x$ $(5x-3)^2dx$ ?

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Answer 1 · 2018-05-25T06:44:33

$\int {(5 x - 3)}^{2} d x = \frac{1}{15} \cdot {(5 x - 3)}^{3} + C$ $int (5x-3)^2dx=1/15*(5x-3)^3+C$

Explanation:

Making the substitution
$t = 5 x - 3$ $t=5x-3$ then $d x = \frac{1}{5} d t$ $dx=1/5dt$
then we have
$\frac{1}{5} \cdot \int t^{2} d t$ $1/5*int t^2dt$
using that $\int x^{n} d x = \frac{x^{n + 1}}{n + 1} + C$ $int x^ndx=x^(n+1)/(n+1)+C$ with $n \neq - 1$ $n ne -1$ then we obatin
$\int {(5 x - 3)}^{2} d x = \frac{1}{15} \cdot {(5 x - 3)}^{3} + C$ $int (5x-3)^2dx=1/15*(5x-3)^3+C$

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How do you integrate ${(5 x - 3)}^{2} d x$ $(5x-3)^2dx$ ?

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How do you integrate ${(5 x - 3)}^{2} d x$ $(5x-3)^2dx$ ?