How to integrate 1/(x^(3/2)+4) ?
1 Answer
Use the substitution
Explanation:
Let
#I=int1/(x^(3/2)+4)dx#
Apply the substitution
#I=int1/(4u^3+4)(2^(7/3)udu)#
Simplify:
#I=2^(1/3)intu/(u^3+1)du#
Factorize the denominator:
#I=2^(1/3)intu/((u^2-u+1)(u+1))du#
Apply partial fraction decomposition:
#I=2^(1/3)/3int((u+1)/(u^2-u+1)-1/(u+1))du#
Rearrange:
#I=2^(1/3)/6int(2u+2)/(u^2-u+1)du-2^(1/3)/3int1/(u+1)du#
For the first term, pull out the a numerator that is the derivative of the denominator:
#I=2^(1/3)/6int((2u-1)/(u^2-u+1)-3/(u^2-u+1))du-2^(1/3)/3ln|u+1|#
Hence
#I=2^(1/3)/6int(2u-1)/(u^2-u+1)du-2^(1/3)/2int1/(u^2-u+1)du-2^(1/3)/3ln|u+1|#
Complete the square in the remaining term:
#I=2^(1/3)/6ln|u^2-u+1|-2^(1/3)int2/((2u-1)^2+3)du-2^(1/3)/3ln|u+1|#
Apply the substitution
#I=2^(1/3)/6ln|u^2-u+1|-2^(1/3)/sqrt3intd theta-2^(1/3)/3ln|u+1|#
Hence
#I=2^(1/3)/6ln|u^2-u+1|-2^(1/3)/sqrt3theta-2^(1/3)/3ln|u+1|+C#
Reverse the last substitution:
#I=2^(1/3)/6ln|u^2-u+1|-2^(1/3)/sqrt3tan^(-1)((2u-1)/sqrt3)-2^(1/3)/3ln|u+1|+C#
Rewrite in terms of
#I=2^(1/3)/6ln|(2^(2/3)u)^2-2^(2/3)(2^(2/3)u)+2^(4/3)|-2^(1/3)/sqrt3tan^(-1)((2^(1/3)(2^(2/3)u)-1)/sqrt3)-2^(1/3)/3ln|2^(2/3)u+2^(2/3)|+C#
Reverse the first substitution:
#I=2^(1/3)/6ln|x-2^(2/3)sqrtx+2^(4/3)|-2^(1/3)/sqrt3tan^(-1)((2^(1/3)sqrtx-1)/sqrt3)-2^(1/3)/3ln|sqrtx+2^(2/3)|+C#