show below
we will solve it by Trigonometric compensation
#intdx/(x²+1)^³#
suppose:
#a^2=1and a=1#
#b^2=1and b=1#
#x=a/btan(theta)=tan(theta)#
#theta=tan^-1(x)#
#dx=sec^2(theta)*d(theta)#
now the integral become:
#int[sec^2(theta)*d(theta)]/(sec(theta)^2)^3=int[sec^2(theta)*d(theta)]/(sec(theta)^6)#
#int[d(theta)]/(sec(theta)^4)=intcos^4(theta)*d(theta)#
note that
#sectheta=1/costheta#
#cos^2(theta)=1/2*(1+cos2(theta))#
#int(1/2*(1+cos2(theta)))^2*d(theta)#
#int1/4+cos2(theta)+1/4*cos^2(2theta)*d(theta)=int1/4d(theta)+intcos2(theta)d(theta)+int1/4*cos^2(2theta)*d(theta)#
#int1/4d(theta)+intcos2(theta)d(theta)+int1/4*[1/2*(1+cos2(theta))]*d(theta)#
#=[1/4theta+sin(2*theta)/2+(cos(theta)*sin(theta)+theta)/8]+c#
#=[1/4tan^-1(x)+sin(2*tan^-1(x))/2+(cos(tan^-1(x))*sin(tan^-1(x))+tan^-1(x))/8]+c#
not that from Pythagorean theorem
#theta=tan^-1(x)#=tantheta=x/1#
#sin(tan^-1(x))=(opposite)/(hypotenuse)=x/sqrt(x+1)#
#sin(2*tan^-1(x))=(2x)/sqrt(x+1)#
#cos(tan^-1(x))=(adjacent)/(hypotenuse)=1/sqrt(x+1)#
#=[[1/4tan^-1(x)]+[[(2x)/sqrt(x+1)]/2]+[(1/sqrt(x+1))*(x/sqrt(x+1))]/8+[(tan^-1(x))/8]+c#
