Problem to solve
Find the indefinite integral \[\int\sqrt{\frac{x+3}{1-x}} dx\]
Solution
STEP 1
Take the integral:
 integral sqrt((x + 3)/(1 - x)) dx
STEP 2
For the integrand sqrt((x + 3)/(1 - x)), substitute u = (x + 3)/(1 - x) and du = (1/(1 - x) - (-x - 3)/(1 - x)^2) dx:
 = 4 integral sqrt(u)/(u + 1)^2 du
STEP 3
For the integrand sqrt(u)/(u + 1)^2, substitute s = sqrt(u) and ds = 1/(2 sqrt(u)) du:
 = 8 integral s^2/(s^2 + 1)^2 ds
STEP 4
For the integrand s^2/(s^2 + 1)^2, use partial fractions:
 = 8 integral1/(s^2 + 1) - 1/(s^2 + 1)^2 ds
STEP 5
Integrate the sum term by term and factor out constants:
 = 8 integral1/(s^2 + 1) ds - 8 integral1/(s^2 + 1)^2 ds
STEP 6
The integral of 1/(s^2 + 1) is tan^(-1)(s):
 = 8 tan^(-1)(s) - 8 integral1/(s^2 + 1)^2 ds
STEP 7
For the integrand 1/(s^2 + 1)^2, substitute s = tan(p) and ds = sec^2(p) dp. Then (s^2 + 1)^2 = (tan^2(p) + 1)^2 = sec^4(p) and p = tan^(-1)(s):
 = 8 tan^(-1)(s) - 8 integral cos^2(p) dp
STEP 8
Write cos^2(p) as 1/2 cos(2 p) + 1/2:
 = 8 tan^(-1)(s) - 8 integral(1/2 cos(2 p) + 1/2) dp
STEP 9
Integrate the sum term by term and factor out constants:
 = 8 tan^(-1)(s) - 4 integral cos(2 p) dp - 4 integral1 dp
STEP 10
For the integrand cos(2 p), substitute w = 2 p and dw = 2 dp:
 = 8 tan^(-1)(s) - 2 integral cos(w) dw - 4 integral1 dp
STEP 11
The integral of cos(w) is sin(w):
 = 8 tan^(-1)(s) - 2 sin(w) - 4 integral1 dp
STEP 12
The integral of 1 is p:
 = -4 p + 8 tan^(-1)(s) - 2 sin(w) + constant
STEP 13
Substitute back for w = 2 p:
 = -4 p - 2 sin(2 p) + 8 tan^(-1)(s) + constant
STEP 14
Substitute back for p = tan^(-1)(s):
 = 4 tan^(-1)(s) - (4 s)/(s^2 + 1) + constant
STEP 15
Substitute back for s = sqrt(u):
 = 4 tan^(-1)(sqrt(u)) - (4 sqrt(u))/(u + 1) + constant
STEP 16
Substitute back for u = (x + 3)/(1 - x):
 = sqrt((x + 3)/(1 - x)) (x - 1) + 4 tan^(-1)(sqrt((x + 3)/(1 - x))) + constant
STEP 17
Which is equivalent for restricted x values to:
Answer: | 
 | = (sqrt((x + 3)/(1 - x)) ((x - 1) sqrt(x + 3) - 4 sqrt(1 - x) sin^(-1)(sqrt(1 - x)/2)))/sqrt(x + 3) + constant
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