Question

$\int\limits^2_1 1/x(1+x²) dx ,Evaluate the given integral expression:$

Answer 1

HELLO DEAR,



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

now, by partial fraction,
1/x(1 + x²) = (Ax + B)/(1 + x²) + C/x

1 = Ax² + Bx + C + Cx²

1 = x²(A + C) + Bx + c

C = 1 , B = 0 , A + C = 0 => A = -1

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

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

put (x² + 1) = t => 2x.dx = dt
also, [x = 2 , t = 5] and [x = 1 , t = 3]

so, $\sf{-1/2\int\limits^5_3 dx/t + [log|x|]^{^2}_1}$

= $\sf{-1/2[log|t|]^{^5}_3 + [log|2| - log|1|]}$

= $\sf{-1/2[log|5| - log|3|] + log|2|]}$

= $\sf{log2 - log3 - 1/2log5}$


I HOPE ITS HELP YOU DEAR,
THANKS