Mathematics integral problem?

Question

Please integrate step by step:

-2 INT (x^(1/2))/(1+x^(1/2)) dx

Thank you in advance.

Answer 1

There is no need to state what is constant, nor any need to change the variable used for the constant as the above answerer has done.



Integrate the original integrand by substitution:

-2 ∫ √x / (1 + √x) dx

Let u = √x,

x = u²

dx / du = 2u

dx = 2u du

-2 ∫ √x / (1 + √x) dx = -4 ∫ u² / (1 + u) du



Divide this expression by the comparing coefficients method:

u² / (1 + u) = Au + B + C / (1 + u)

u² = Au(1 + u) + B(1 + u) + C

u² = Au + Au² + B + Bu + C

u² = Au² + (A + B)u + (B + C)

A = 1

A + B = 0

B = -A

B = -1

B + C = 0

C = -B

C = 1

u² / (1 + u) = u - 1 + 1 / (1 + u)



Integrate the expression term by term using this result:

∫ u² / (1 + u) du = ∫ [u - 1 + 1 / (1 + u)] du

∫ u² / (1 + u) du = ∫ u du - ∫ 1 du + ∫ 1 / (1 + u) du

∫ u² / (1 + u) du = u² / 2 - u + ln|1 + u|



Put it all together to integrate the original function:

-2 ∫ √x / (1 + √x) dx = -4 ∫ u² / (1 + u) du

-2 ∫ √x / (1 + √x) dx = -4[u² / 2 - u + ln|1 + u|] + C

-2 ∫ √x / (1 + √x) dx = -2u² + 4u - 4ln|1 + u| + C

-2 ∫ √x / (1 + √x) dx = 4u - 2u² - 4ln|u + 1| + C

Since u = √x,

-2 ∫ √x / (1 + √x) dx = 4√x - 2x - 4ln(√x + 1) + C

Answer 2

I will concentrate only on the integral. The coefficients in front can be worried about later.

Define: I = INT (x^(1/2))/(1+x^(1/2)) dx

Put (x^(1/2)) = t

=> x = (t^2)

=> dx = 2*t*dt



So I now becomes:

I = INT (t/(1+t))*(2*t*dt)

I = 2* INT (t^2/(1+t))*dt



Now, J = INT (t^2/(1+t))*dt

J = INT ((t^2-1+1)/(1+t))*dt

J = INT((t^2-1)/(1+t))*dt + INT (1/(1+t))*dt

J = INT({(t-1)*(t+1)}/(1+t))*dt + INT (1/(1+t))*dt

J = INT((t-1)*dt) + INT(1/(1+t))*dt

J = ((t-1)^2)/2 + ln(1+t) + C, C = arbitrary constant



Now substitute back for t

J = (([x^(1/2)]-1)^2)/2 + ln(1+[x^(1/2)]) + C

I = 2*(([x^(1/2)]-1)^2)/2 + ln(1+[x^(1/2)]) + C2, C2 = arbitrary constant

Answer 3

For INT (x^(1/2))/(1+x^(1/2)) dx

let u=1+x^(1/2). Then

du=1/2x^(-1/2)dx=> 2xdu=2(u-1)du=dx and the intergral becomes

int{2(u-1)^2/udu}=int{2(u-2+1/u)du}

=2int{u-2+1/u}du

=2*(u^2/2-2u+Log(u)}

=(x^1/2+1)^2-4(x^(1/2)+1)+2log(x^(1/2)+1)

=x-2x^(1/2)+2log(x^(1/2)+1)-3.

Absorbing the -3 into the constant of integration gives

INT (x^(1/2))/(1+x^(1/2)) dx=x-2sqrt(x)+2log(sqrt(x)+1)+C



Hope that helps.

Answer 4

-2*(-2*Sqrt[x] + x + 2*Log[1 + Sqrt[x]]) +c :x>= 0