Asked by Jay
                Trying to find ∫x*arctan(x)dx, but I can't figure out what do to after:
(1/2)x^2*arctan(x)-1/2∫x^2/(x^2+1) dx
            
        (1/2)x^2*arctan(x)-1/2∫x^2/(x^2+1) dx
Answers
                    Answered by
            Bosnian
            
    In google type: 
wolfram alpha
When you see list of results click on:
Wolfram Alpha:Computational Knowledge Engine
When page be open in rectangle type:
integrate x*arctan(x)dx
and click option =
After few secons you will see result.
Then click option : Show steps
    
wolfram alpha
When you see list of results click on:
Wolfram Alpha:Computational Knowledge Engine
When page be open in rectangle type:
integrate x*arctan(x)dx
and click option =
After few secons you will see result.
Then click option : Show steps
                    Answered by
            Jay
            
    great, I can find the steps and i have the answer already. i just don't understand the step to be made after (1/2)x^2*arctan(x)-1/2∫x^2/(x^2+1) dx
    
                    Answered by
            Ethan
            
    To solve that last integral, you have to use the substitution rule.
u= x^2 + 1
du= 2x
Therefore the integral becomes:
-1/2∫(2/u) du
Factor out the 2
-∫1/u du
Integrate
-ln(u)
Which is equal to
-ln(x^2 +1)
Since u= x^2 + 1
    
u= x^2 + 1
du= 2x
Therefore the integral becomes:
-1/2∫(2/u) du
Factor out the 2
-∫1/u du
Integrate
-ln(u)
Which is equal to
-ln(x^2 +1)
Since u= x^2 + 1
                    Answered by
            Jay
            
    I figured it out
@Ethan, you can't substitute because it is x^2 over(x^2+1)
    
@Ethan, you can't substitute because it is x^2 over(x^2+1)
                    Answered by
            MathMate
            
    x^2/(x^2+1) = 1 - 1/(x^2+1)
You can easily integrate that.
    
You can easily integrate that.
                    Answered by
            Ethan
            
    Sorry, I misread that as 2x instead of x^2.
    
                    Answered by
            MathMate
            
    No problem!
    
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