Question

$\int\limits^3_1 dx/x²(x+1) ,Evaluate the given integral expression:$

Answer 1

HELLO DEAR,



GIVEN:-
$\sf{\int\limits^3_1 dx/x^2(x + 1)}$

NOW,

solving, 1/x²(x + 1) = A/(1 + x) + (Bx + C)/x²

1 = Ax² + (1 + x)(Bx + C)

1 = Ax² + Bx² + Bx + Cx + C

1 = x²(A + B) + x(B + C) + C

therefore, C = 1 , A + B = 0 , B + C = 0

B = -1 , A = 1

so, $\sf{\int\limits^3_1 [1/(1 + x) + (1 - x)/x^2]\,dx}$

$\sf{= [log|1 + x|]^{^3}_1 + \int\limits^3_1[1/x^2 - 1/x]\,dx}$

$\sf{= log|4| - log|2| + [-1/x - log|x|]^{^3}_1}$

$\sf{= 2log2 - log2 + [-1/3 + 1 - log3 + log1]}$

$\sf{=log2 + 2/3 - log3}$


I HOPE ITS HELP YOU DEAR,
THANKS