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

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

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Answer

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

Step by step solution

Problem

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

The integral of a constant is equal to the constant times the integral's variable

$x\frac{v}{1-2v+v^2}$
2

Multiplying the fraction and term

$\frac{v\cdot x}{1-2v+v^2}$
3

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

$\Delta=b^2-4ac=-2^2-4\left(1\right)\left(1\right) = 0$
4

Using the perfect square trinomial formula

$a^2+2ab+b^2=(a+b)^2,\:where\:a=\sqrt{v^2}\:and\:b=\sqrt{1}$
5

Factoring the perfect square trinomial

$\frac{v\cdot x}{\left(v-1\right)^{2}}$
6

Add the constant of integration

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

Answer

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

Problem Analysis

Main topic:

Integral calculus

Time to solve it:

0.27 seconds

Views:

116