How do you find the integral #intx^3/(sqrt(x^2+9))dx# ?
1 Answer
Aug 29, 2014
#=(x^2+9)^(3/2)/3-9sqrt(x^2+9)+c# , where#c# is a constantExplanation :
#=intx^3/sqrt(x^2+9)dx# Using Integration by Substitution,
Let's assume
#x^2+9=t^2# , then
#2xdx=2tdt# ,#=>xdx=tdt#
#=int((t^2-9)t)/tdt#
#=intt^2dt-int9dt#
#=t^3/3-9t+c# , where#c# is a constantSubstituting
#t# back,
#=(x^2+9)^(3/2)/3-9sqrt(x^2+9)+c# , where#c# is a constant
