What is the indefinite integral of 1 + x ^ 2 under the root sign? It is necessary to explain the process or method

What is the indefinite integral of 1 + x ^ 2 under the root sign? It is necessary to explain the process or method

Using the second integral transformation method, let x = TANU, u ∈ (- π / 2, π / 2), then ∫ √ (1 + x) ²） dx=∫sec ³ udu=∫secudtanu=secutanu-∫tanudsecu=secutanu-∫tan ² usecudu=secutanu-∫sec ³ udu+∫secudu=secutanu+1/2ln|secu+tan...

Indefinite integral arctan radical x DX

Step integration method
Original formula = xarctan √ X - ∫ xdarctan √ x
=xarctan√x-∫x/(1+x)dx
=xarctan√x-∫(x+1-1)/(1+x)dx
=xarctan√x-∫[1-1/(1+x)]dx
=xarctan√x-x+ln(1+x)+C

Solving indefinite integral: under the root sign [(1-x) / (1 + x)] DX

∫√[(1-x)/（1+x)]dx
=∫(1-x)/√(1-x^2)dx
=∫1/√(1-x^2)-∫x/√(1-x^2)dx
=arcsinx+1/2∫(1-x^2)^(-1/2)d(1-x^2)
=arcsinx+√(1-x^2)+c
C is a constant

Calculate the indefinite integral, the integral number arctan (x under the root sign) DX

∫ arctan(√x) dx
Partial integral
=xarctan(√x) - ∫ x/(1+x) d(√x)
=xarctan(√x) - ∫ (x+1-1)/(1+x) d(√x)
=xarctan(√x) - ∫ 1 d(√x) + ∫ 1/(1+x) d(√x)
=xarctan(√x) - √x + arctan(√x) + C
If you don't understand, please ask. If you solve the problem, please click "select as satisfactory answer" below

Find the indefinite integral of ∫ [arctan √ X / √ (1 + x)] DX,

t = arctan√x ,sect = √(1+x),x = tan ² t ,dx = 2 tan t * sec ² T DT original formula = ∫ 2 T d (sect) = 2 T * sect - 2 ∫ sect DT = 2 T * sect - 2 ln| sect + tant| + C = 2 √ (1 + x) arctan √ X - 2 ln| √ (1 + x) + √ X

Find indefinite integral (1 / radical x (1 + x)) DX

Match the formula in the root sign to (x + 1 / 2) ^ 2-1 / 4 under the root sign
That is (x + 1 / 2) ^ 2 - (1 / 2) ^ 2
Conform to the formula of an indefinite integral (or change the element, let t = x + 1 / 2)
The result is x (1 + x) | + C under the root sign of in|x + 1 / 2 +