what is the integral of?

Question

what is the integral of sqrt((a+x)/(b+x))?

Answer 1

Int sqrt((a+x)/(b+x)) dx

Let us add and subtract “b” in the numerator
Int [(a-b)+(x+b)} / (b+x) ]^1/2 *dx ----(1)

let b+x = (a-b) tan^2 (T) so
dx = 2(a-b) tan (T)* sec^2 (T). dT put in (1)

= Int 2(a-b) tan (T)*sec^2 (T).dT [sec^2 (T) / tan^2 (T)]^1/2

= Int 2(a-b) tan (T)* sec^2 (T). dT [sec (T) / tan (T)]

I = Int 2(a-b) sec^3 (T) dT or

I / 2(a-b) = Int sec^3 (T) dT = J --------- (2)

The integral of sec^3 (T) has to be calculated separately let it be “ J “

J = Int sec^3 (T) dT = Int {sec (T) } {sec^2 (T) dT }

(2) do it by parts integ {u}. dv
Where u=sec (T), dv = sec^2 (T) dT or v = tan (T)

J= {sec (T)*tan (T) – Int {tan (T)*sec (T) tan (T) dT }
={sec (T)*tan (T) – Int {sec (T) (Sec^2 (T) -1) dT
={sec (T)*tan (T) – J + Int {sec (T) dT
2J = {sec (T)*tan (T) + log {tan (T) + sec (T)}
J = (1/2){sec (T)*tan (T) + log {tan (T) + sec (T)} put in (2)

I= 2(a-b) J = (a-b) [sec (T)*tan (T) + log {tan (T) + sec (T)}]

Integrated = (a-b)[sec (T)*tan (T) + log {tan (T) + sec (T)}]----(3)

Changing T into x form from
b+x = (a-b) tan^2 (T) also a+x = (a-b) sec^2 (T)
(a-b) sec (T)*tan (T) = sqrt {(a+x)(b+x)} ---- (4)
sec (T) + tan (T) = {sqrt (a+x) + sqrt (b+x)} / sqrt (a-b) ---(5)

put (4) and (5) in (3) final answer

=sqrt{(a+x)(b+x)}+ (a-b) log[{sqrt (a+x) + sqrt (b+x)} / sqrt (a-b)]-------- (6)

Answer 2

{Disclaimer: this may or may not be the most efficient way to do this; nevertheless I'll be using valid mathematical steps.}

Let u = sqrt(a + x). Then
u^2 = a + x

u^2 - a = x

u^2 - a + b = b + x, so
2u du = dx, and we have

Integral [ (u/(u^2 - a + b)) 2u du ]

2 * Integral [u^2/(u^2 - a + b) du]

Using long division on u^2 and u^2 - a + b will yield

u^2 /(u^2 - a + b) = (-a + b)/(u^2 - a + b) + 1

So we have

2 * Integral [(-a + b)/(u^2 - a + b) + 1] du

2 * { Integral [(-a + b)/(u^2 - a + b)] du + Integral(1)du }

2 * { Integral [(-a + b)/(u^2 - a + b)] du + u }

2 * {(-a + b) Integral [ 1/(u^2 - a + b) du ] + u } + C

And this has become a trig substitution problem.

Let u = sqrt(-a + b)sec(t). Then
du = sqrt(-a + b)sec(t)tan(t) dt.

(skipping details)
Integral [ 1/(u^2 - a + b) du ] becomes

sqrt(-a + b)/(-a + b) * Integral [ sec(t) dt ], which becomes
sqrt(-a + b)/(-a + b) * ln|sec(t) + tan(t)|

But, if u = sqrt(-a + b)sec(t), then
sec(t) = u/sqrt(-a + b) = hyp/adj
hyp = u
adj = sqrt(-a + b). Therefore, by Pythagoras,
opp = sqrt[u^2 - (-a + b)]

tan(t) = opp/adj = sqrt[u^2 - (-a + b)]/sqrt(-a + b)

2 * {(-a + b) sqrt[u^2 - (-a + b)]/sqrt(-a + b) + u } + C

2 * {sqrt (-a + b) sqrt[u^2 - (-a + b)] + u } + C

2sqrt (-a + b) sqrt[u^2 - (-a + b)] + 2u + C

But u = sqrt(a + x), so

2sqrt (-a + b) sqrt[(a + x) - (-a + b)] + 2sqrt(a + x) + C

2sqrt(-a + b) sqrt(x + 2a + - b) + 2sqrt(a + x) + C

I cannot verify the validity of this answer.

Answer 3

(a+x)/(b+x). The integral is the component squared.