#int x^2 / (x^2 + 4) dx# integrate?

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Answer 1 · 2018-04-14T09:20:34

#int x^2 / (x^2 + 4) dx= x -2 tan^-1(x/2) + C#

Given: #int x^2 / (x^2 + 4) dx#

Add 0 to the numerator in the form of #+4-4#:

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

Separate into two fractions:

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

The first fraction becomes 1:

#int x^2 / (x^2 + 4) dx= int 1-4 / (x^2 + 4) dx#

Separate into two integrals:

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

The first integral is trivial and the second integral is the inverse tangent:

#int x^2 / (x^2 + 4) dx= x -2 tan^-1(x/2) + C#