How to integrate 1/(x^(3/2)+4) ?

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Answer 1 · 2018-06-08T00:53:21

Use the substitution $x^(1/2)=2^(2/3)u$ .

Let

$I=int1/(x^(3/2)+4)dx$

Apply the substitution $x^(1/2)=2^(2/3)u$ :

$I=int1/(4u^3+4)(2^(7/3)udu)$

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 $2u-1=sqrt3tantheta$ :

$I=2^(1/3)/6ln|u^2-u+1|-2^(1/3)/sqrt3intd theta-2^(1/3)/3ln|u+1|$

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 $2^(2/3)u$ :

$I=2^(1/3)/6ln|(2^(2/3)u)^2-2^(2/3)(2^(2/3)u)+2^(4/3)|-2^(1/3)/sqrt3tan^(-1)((2^(1/3)(2^(2/3)u)-1)/sqrt3)-2^(1/3)/3ln|2^(2/3)u+2^(2/3)|+C$

Reverse the first substitution:

$I=2^(1/3)/6ln|x-2^(2/3)sqrtx+2^(4/3)|-2^(1/3)/sqrt3tan^(-1)((2^(1/3)sqrtx-1)/sqrt3)-2^(1/3)/3ln|sqrtx+2^(2/3)|+C$