Question

Q from Definite Integrals​

Answer 1

Step-by-step explanation:

We have,

$i = 5 \int^{2} _{1} \frac{ {x}^{2} + 4x + 3 - 4x - 3}{ {x}^{2} + 4x + 3} dx$

$i= 5 \int^{2} _{1}(1 - \frac{4x + 3}{ {x}^{2} + 3x + x + 3 } )dx$

$i = 5\int^{2} _{1}dx - 5 \int^{2} _{1} \frac{4x + 3}{(x + 3)(x + 1)}dx$

$i= 5 [x]_{1}^{2} - 5 \int^{2} _{1} \frac{4x + 3}{(x + 3)(x + 1)} dx$

$i = 5 - 5 \int^{2} _{1} \frac{4x + 3}{(x + 3)(x + 1)} dx$

$let \: \frac{4x + 3}{(x + 3)(x + 1)} = \frac{a}{x + 3} + \frac{b}{x + 1}$

$\implies \frac{4x + 3}{(x + 3)(x + 1)} = \frac{(a + b)x + (a + 3b)}{(x + 3)(x + 1)}$

$\implies \: a + b = 4 \: \: and \: \: a + 3b = 3$

Solving these , we get,

$\implies \: a = \frac{9}{2} \: \: and \: \: b = - \frac{1}{2}$

Now,

$i = 5 - 5 \int^{2} _{1} \frac{9}{2(x + 3)} dx - 5 \int^{2} _{1} \frac{ - 1}{2(x + 1)} dx$

$i = 5 - \frac{45}{2} \int^{2} _{1} \frac{dx}{x + 3} + \frac{5}{2} \int^{2} _{1} \frac{dx}{x + 1}$

$i = 5 - \frac{45}{2} [ ln(x + 3) ] ^{2} _{1} - \frac{5}{2} [ ln(x + 1) ]^{2} _{1}$

$i = 5 - \frac{45}{2} [ 2ln(2) - ln(3) ] + \frac{5}{2} [ ln(3) - ln(2) ]$

$i =5 - {45} ln(2) + \frac{45}{2} ln(3) + \frac{5}{2} ln(3) - \frac{5}{2} ln(2)$

$i = 5 - 25 ln(2) + 25 ln(3) - \frac{45}{2} ln(2)$

$i = 5 + 25 ln( \frac{3}{2} ) - \frac{45}{2} ln(2)$