Asked by Sean
                Integral of (dt/[sqrt(t)+25tsqrt(t)]}
From 1/75 to 3/25
I'm at a complete loss at what to do.
            
        From 1/75 to 3/25
I'm at a complete loss at what to do.
Answers
                    Answered by
            Damon
            
    let y = t^(1/2)
dy = (1/2) t^-(1/2) dt
dt = 2 t^(1/2) dy = 2 y dy
sqrt t = y
25 t sqrt t = 25 y^3
so
integral 2 y dy/(y+25 y^3)
or
integral of 2 dy/(1 + 25 y^2)
    
dy = (1/2) t^-(1/2) dt
dt = 2 t^(1/2) dy = 2 y dy
sqrt t = y
25 t sqrt t = 25 y^3
so
integral 2 y dy/(y+25 y^3)
or
integral of 2 dy/(1 + 25 y^2)
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