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 , wherec 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 , wherec is a constantSubstituting
t back,
=(x^2+9)^(3/2)/3-9sqrt(x^2+9)+c , wherec is a constant