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

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Answer 1 · 2016-02-07T12:05:20

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

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#