How do you integrate $\frac{7}{{(x - 3)}^{2}}$ $7/(x-3)^2$ using partial fractions?

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How do you integrate $\frac{7}{{(x - 3)}^{2}}$ $7/(x-3)^2$ using partial fractions?

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Answer 1 · 2017-10-12T21:56:26

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

Explanation:

$\frac{7}{{(x - 3)}^{2}} = (\frac{A 1}{x - 3}) + (\frac{A 2}{{(x - 3)}^{2}})$ $7/(x-3)^2=((A1)/(x-3))+((A2)/(x-3)^2)$
$A 1 \cdot (x - 3) + A 2 = 7$ $A1*(x-3)+A2=7$
For x = 3,
$A 2 = 7$ $A2=7$
Comparing coefficient of x, we get $A = 0$ $A=0$
Let $t = (x - 3), d t = d x$ $t=(x-3), dt=dx$
$\int (\frac{7}{t^{2}}) d t = (- \frac{7}{t}) + c$ $int(7/t^2)dt=(-7/t)+c$
Substituting value of $t = (x - 3)$ $t=(x-3)$
$= \frac{- 7}{x - 3} + c$ $=(-7)/(x-3)+c$

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How do you integrate $\frac{7}{{(x - 3)}^{2}}$ $7/(x-3)^2$ using partial fractions?

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