Asked by O_o Rion
                Integral of x sq/(1+x sq) dx 
I am not sure how to do this, am i supposed to do partial fractions to break it up? Or is it just some simple substitution?
            
        I am not sure how to do this, am i supposed to do partial fractions to break it up? Or is it just some simple substitution?
Answers
                    Answered by
            Reiny
            
    If I see the derivative of the base sitting out front either directly of as a multiple, I usually can do the integral intuitively 
I know it has to be (?)(1 + x^2)^(3/2)
I then visulaize differentiating that, and adjust the coefficient accordingly.
so it should be (1/3)(1 + x^2)^(3/2)
Your teacher probably wants you to do it by substitution
let u = (1 + x^2)
du/dx = 2x
dx = du/2x
so Integral of x sq/(1+x sq) dx
= integral (1/2)u^(1/2) du
= (1/3)u^(3/2)
= (1/3)(1 + x^2)^(3/2)
    
I know it has to be (?)(1 + x^2)^(3/2)
I then visulaize differentiating that, and adjust the coefficient accordingly.
so it should be (1/3)(1 + x^2)^(3/2)
Your teacher probably wants you to do it by substitution
let u = (1 + x^2)
du/dx = 2x
dx = du/2x
so Integral of x sq/(1+x sq) dx
= integral (1/2)u^(1/2) du
= (1/3)u^(3/2)
= (1/3)(1 + x^2)^(3/2)
                    Answered by
            O_o Rion
            
    But how did you get
Integral of x sq/(1+x sq) dx
= integral (1/2)u^(1/2) du
du= 2xdx, from there i got,
integral xsq/(1+x) dx
= (1/2) integral[ (sqrt (1-u))/u ]du
Did i do anything wrong?
    
Integral of x sq/(1+x sq) dx
= integral (1/2)u^(1/2) du
du= 2xdx, from there i got,
integral xsq/(1+x) dx
= (1/2) integral[ (sqrt (1-u))/u ]du
Did i do anything wrong?
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