How would I evaluate the integral of 1/(x^2 + a^2) ?

Answer 1

if a is constant.
use trig substitute
x= atan(Q) (1)
dx=asec^2(Q)dQ
put it back in the integral
asec^2(Q)dQ/(a^2tan^2(Q)+a^2)
factor a^2 we have:

asec^2(Q)dQ/(a^2(tan^2(Q)+1))
but tan^2(Q)+1=sec^2(Q)
integral( asec^2(Q)dQ/(a^2sec^2Q))
int(1/a dQ)
1/a*(Q) (2)
from (1)
tanQ=x/a
=>Q=tan^-1(x/a)
but that into(2)
so the answer is
1/a*(tan^-1(x/a) +C
remember !!!! tan^-1 is called
inverse tangent. good luck

Answer 2

Evaluate the integral of 1/(x² + a²). I assume you mean with respect to x and that a is a constant.

∫{1/(x² + a²)}dx
Let
atanθ = x
asec²θ dθ = dx

∫{1/(x² + a²)}dx
= ∫{1/((atanθ)² + a²)}(asec²θ dθ)
= ∫{1/(a²sec²θ)}(asec²θ dθ)
= ∫{1/a}dθ
= θ/a + C
= (1/a)arctan(x/a) + C