How do you find the integral of $(sqrt(1+x^2)/x)$ ?

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Answer 1 · 2015-07-01T17:54:21

The answer is $[1/2ln(1/sqrt(x^2+1)-1)+sqrt(x^2+1)-1/2ln(1/sqrt(x^2+1)+1)]+C$

let's $x = tan(u)$

$dx = 1/cos^2(u)du$

the integral become :

$int (1/cos(u))/tan(u)*1/cos^2(u) du$

$int 1/tan(u)*1/cos^3(u) du$

$int 1/(sin(u)cos^2(u)$

$int sin(u)/sin(u)^2cos^2(u)$

$t = cos(u)$

$dt = -sin(u)$

$-int1/((1-t^2)t^2)$

$-int1/((1+t)(1-t)t^2$

with partial fraction we get

$-int-1/(2 (t-1)) + 1/t^2 + 1/(2 (1 + t))$

$[1/2ln(t-1)+1/t-1/2ln(t+1)]+C$

since $t = cos(u)$

$[1/2ln(cos(u)-1)+1/cos(u)-1/2ln(cos(u)+1)]+C$

remember $x = tan(u)$

so $arctan(x) = u$

$cos(arctan(x)) = cos(u)$

but $cos(arctan(x)) = 1/sqrt(x^2+1)$

FINALLY :

$[1/2ln(1/sqrt(x^2+1)-1)+sqrt(x^2+1)-1/2ln(1/sqrt(x^2+1)+1)]+C$

How do you find the integral of (sqrt(1+x^2)/x)(sqrt(1+x^2)/x)?