Question

integral 2x+1/x2-3x+2 DX can be used partial fraction




​

Answer 1

Solution:-

We have

$\sf \implies \: \int \dfrac{2x + 1}{ {x}^{2} - 3x + 2 } dx$

Now Take

$\implies \sf \: {x}^{2} - 3x + 2$

When we factories the polynomial we get

$\implies \sf \: {x}^{2} - 3x + 2 = (x - 1)(x - 2)$

we can write as

$\sf \implies \: \int \dfrac{2x + 1}{(x - 1)(x - 2)}dx$

Now using partial Fraction method, So Take

$\sf\implies \: \dfrac{2x + 1}{(x - 1)(x - 2)} \: \: = \dfrac{A}{(x - 1)} + \dfrac{B}{(x - 2)}$

Now Take LCM

$\sf \implies \: \dfrac{2x + 1}{(x - 1)(x - 2)} = \dfrac{A(x - 2) +B(x - 1) }{(x - 1)(x - 2)}$

$\sf \implies \: \dfrac{2x + 1}{ \cancel{(x - 1)(x - 2)}} = \dfrac{A(x - 2) +B(x - 1) }{ \cancel{(x - 1)(x - 2)}}$

$\sf \implies \: 2x + 1 = A(x - 2) + B(x - 1)$

Now when x = 2

$\sf \implies \: 2 \times 2+ 1 = A(2 - 2) + B(2 - 1)$

$\sf \implies \: 5 = 0 + B \implies \: B = 5$

Now when x = 1

$\sf \implies \: 2 \times 1 + 1 = A(1 - 2) + 0$

$\sf \implies \: 3 = - A \implies \:A = - 3$

Now put the value of A and B on

$\sf\implies \: \dfrac{2x + 1}{(x - 1)(x - 2)} \: \: = \dfrac{ - 3}{(x - 1)} + \dfrac{5}{(x - 2)}$

$\sf \implies \int\dfrac{ - 3}{(x - 1)}dx + \int\dfrac{5}{(x - 2)} dx$

$\sf \implies \: - 3 \int \dfrac{1}{x - 1} dx + 5 \int \dfrac{1}{x - 2} dx$

$\sf \implies \: - 3 \ln(x - 1) + 5 \ln(x - 2) + c$

Answer

$\sf \implies \: - 3 \ln(x - 1) + 5 \ln(x - 2) + c$