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Question
evaluate integral
x^3/sqrt(x^2+1) dx
Hello Sunita, let's use the substitue methods :
Let u=√(x^2+1) . (Where "√" resimble square root), Then
u^2=x^2+1 -> x=√(u^2-1). Now let's calculate dx : from our
substitute we get : du=xdx/√(x^2+1) -> dx=(du/x)√(x^2+1) ->
dx=udu/√(u^2-1). & it's easy to see that x^3=(u^2-1)^(3/2).
Now our integral ∫x^3/√(x^2+1) dx will become :
∫u/√(u^2-1)*(u^2-1)^(3/2) /u du
Which is also equal to ∫(u^2-1)^(3/2) / √(u^2-1) du
=∫(u^2-1)du = u^3/3 - u . Now let's substitute u=√(x^2+1) :
=(1/3)[√(x^2+1)]^3-√(x^2+1)
=(1/3)√(x^2+1)[x^2+1-3]
=(1/3)(x^2-2)√(x^2+1)
Alon.
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Alon Mandes
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Kind of questions I can answer : Limits, Derivatives, Integration, Implicit functions, continuousity, differentiation ,Extremum problems, Lagrange multipliers, Gradients, Surface integrals, Multi variables functions ,Multi variables Integrals,Complex variables ,Complex functions, Curves, Trajectory integrals & Vector analyse,Divergence,Rotor & word problems. Kind of question I can't answer : Economics,Combinatorics,infinite series & convergence ,Statistics & Probabilities .
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