# Integral of v/(-2v+1+v^2)

## \int\frac{v}{v^2-2v+1}dx

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$\frac{v\cdot x}{\left(v-1\right)^{2}}+C_0$

## Step by step solution

Problem

$\int\frac{v}{v^2-2v+1}dx$
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The integral of a constant is equal to the constant times the integral's variable

$x\frac{v}{1-2v+v^2}$
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Multiplying the fraction and term

$\frac{v\cdot x}{1-2v+v^2}$
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The trinomial $\frac{v\cdot x}{1-2v+v^2}$ is perfect square, because it's discriminant is equal to zero

$\Delta=b^2-4ac=-2^2-4\left(1\right)\left(1\right) = 0$
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Using the perfect square trinomial formula

$a^2+2ab+b^2=(a+b)^2,\:where\:a=\sqrt{v^2}\:and\:b=\sqrt{1}$
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Factoring the perfect square trinomial

$\frac{v\cdot x}{\left(v-1\right)^{2}}$
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$\frac{v\cdot x}{\left(v-1\right)^{2}}+C_0$

$\frac{v\cdot x}{\left(v-1\right)^{2}}+C_0$

### Main topic:

Integral calculus

0.27 seconds

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