How do you find the integral intx^2/((x^2+2)^(3/2))dx ?
1 Answer
Aug 29, 2014
=1/2*x/sqrt(2+x^2)+c , wherec is a constantExplanation :
=intx^2/((x^2+2)^(3/2))dx
=intx^2/(x^3(1+2/x^2)^(3/2))dx
=int1/(x(1+2/x^2)^(3/2))dx Using Integration by Substitution,
let's assume
2/x^2=t then
-4/xdx=dt
=int-dt/(4(1+t)^(3/2))
=-1/4int(1+t)^(-3/2)dt
=-1/4*((1+t)^(-1/2))/(-1/2)
=1/2*1/sqrt(1+t)+c , wherec is a constantSubstituting
t back,
=1/2*1/sqrt(1+2/x^2)+c , wherec is a constant
=1/2*x/sqrt(2+x^2)+c , wherec is a constant
