Question

Question #13763

Answer 1

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

Explanation:

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

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

Explanation:

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`