How do find #int (x^2) /( x^2 + 4) dx # ?

1 Answer
Jan 7, 2018

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

Explanation:

#intx^2/(x^2+4)dx#

#=int(x^2+4-4)/(x^2+4)dx#

#=int(x^2+4)/(x^2+4)-4/(x^2+4)dx#

#=int1-4/(x^2+4)dx = intdx - int4/(x^2+4)dx#

For the second integral substitute: #2tan(u) = x#
#-> 2sec^2(u)du = dx#

#=intdx-4int(2sec^2(u))/(4tan^2(u)+4)du#

#=intdx-2int(sec^2(u))/(tan^2(u)+1)du#

Now: #sin^2(u)+cos^2(u)=1#

Divide by #cos^2(u)# to get:

#-> sin^2(u)/cos^2(u)+cos^2(u)/cos^2(u)=1/cos^2(u)#

#-> tan^2(u)+1=sec^2(u)#

We can simplify our integral to:

#intdx-2intsec^2(u)/sec^2(u)du#

#=intdx-2intdu=x-2u+C#

Reverse the substitution for #u# to get:

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

Edit: Of course if you knew the integral at the step before I did the substitution, i.e. that:

#int1/(x^2+a^2)dx =1/a tan^(-1)(x/a)#

then you can just integrate directly rather than using substitution.