Question #13763

Question

1159 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2017-06-18T15:02:52

$1/atan^-1(x/a) + "constant"$

Factor out an $a^2$

$int 1/(x^2+a^2)dx=int 1/(a^2((x^2)/(a^2)+1))dx$

Factor out a constant

$=1/a^2int1/((x^2)/(a^2)+1)dx$

Let $u=x/a$ and $du=1/a dx$ , so that $adu=dx$

$=1/a^cancel(2)intcancel(a)/(u^2+1)du$

$=1/aint1/(u^2+1)du$

$=1/a tan^-1(u)+"constant"$

$=1/atan^-1(x/a) + "constant"$

Answer 2 · 2017-06-18T15:08:29

Substitute $x = atan(theta), dx = asec^2(theta)d theta$

Given: $int1/(x^2+a^2)dx =$

Substitute $x = atan(theta), dx = asec^2(theta)d theta$

$int(asec^2(theta))/((atan(theta))^2+a^2)d theta =$

$int(asec^2(theta))/(a^2tan^2(theta)+a^2)d theta =$

$int(sec^2(theta))/(a(tan^2(theta)+1))d theta =$

We know that $sec^2(theta)=(tan^2(theta)+1)$ so the integrand becomes $1/a$ :

$1/aintd theta = 1/atheta+C$

Solve the substitution for $theta$ and then substitute:

$x = atan(theta)$

$x/a = tan(theta)$

$theta = tan^-1(x/a)$

$int1/(x^2+a^2)dx = 1/atan^-1(x/a)+C$