Question

What is the antiderivative of `2/(x^2+1)`?

Answer 1

`2tan^-1(x)+C`

Explanation:

Normally this may be quoted as a standard integral:

`int2/(x^2+1)dx = 2int1/(x^2+1)dx=2tan^-1(x)+C`

However to convince yourself this is indeed the case:

First consider the trig - identity:

`sin^2(x)+cos^2(x)=1`

Divide this through by `cos^2(x)` to obtain the identity:

`sin^2(x)/cos^2(x)+cos^2(x)/cos^2(x)=1/cos^2(x) -> tan^2(x)+1 = sec^2(x)`

Now going back to the integral, use the substitution:

`tan(u) =x`

This will also mean:

`sec^2(u)du = dx` Now put this substitution into the integral:

`int2/(x^2+1)dx=2intsec^2(u)/(tan^2(u)+1)du`

Now using the trig-identity we just saw we can replace the denominator giving us:

`2intsec^2(u)/sec^2(u)du=2int1du=u+C`

Now reverse the substitution and we get:

`=2tan^-1(x) +C`