What is the integral of (1-sqrt(x+1))/(1+sqrt(x+1)) ?

The integral of (1-sqrt(x+1))/(1+sqrt(x+1)) is -x+4sqrt(x+1)-4ln|sqrt(x+1)+1|+C

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integral of (1-sqrt(x+1))/(1+sqrt(x+1))

\int \frac{1 - x + 1}{1 + x + 1} d x

- x + 4 x + 1 - 4 ln x + 1 + 1 + C

Solution steps

\int \frac{1 - x + 1}{1 + x + 1} d x

Apply u-substitution

: \int \frac{2 u ( 1 - u )}{1 + u} d u

= \int \frac{2 u ( 1 - u )}{1 + u} d u

Take the constant out:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

= 2 \cdot \int \frac{u ( 1 - u )}{1 + u} d u

Apply u-substitution

: \int \frac{( v - 1 ) ( - v + 2 )}{v} d v

= 2 \cdot \int \frac{( v - 1 ) ( - v + 2 )}{v} d v

Apply Integration By Parts

: (v - 1) (- v + 2) - \int \frac{- v ^{2} + 2}{v} d v

= 2 ((v - 1) (- v + 2) - \int \frac{- v ^{2} + 2}{v} d v)

\int \frac{- v ^{2} + 2}{v} d v = - \frac{v ^{2}}{2} + 2 ln ∣ v ∣

= 2 ((v - 1) (- v + 2) - (- \frac{v ^{2}}{2} + 2 ln ∣ v ∣))

Substitute back

= 2 ((1 + x + 1 - 1) (- (1 + x + 1) + 2) - (- \frac{( 1 + x + 1 ) ^{2}}{2} + 2 ln 1 + x + 1))

Simplify

2 ((1 + x + 1 - 1) (- (1 + x + 1) + 2) - (- \frac{( 1 + x + 1 ) ^{2}}{2} + 2 ln 1 + x + 1)) : - x + 4 x + 1 - 4 ln x + 1 + 1

= - x + 4 x + 1 - 4 ln x + 1 + 1

Add a constant to the solution

= - x + 4 x + 1 - 4 ln x + 1 + 1 + C

x \to 0 lim (x ∣ x ∣)

\frac{e ^{- 2 s}}{s ^{2} ( s - 1 )}

\frac{\partial}{\partial x} (e^{- 2 x} \cdot cos (2 π y))

\frac{d}{d x} ((sec^{2} (x)) d x)

\frac{d y}{d x} + y = 9

What is the integral of (1-sqrt(x+1))/(1+sqrt(x+1)) ?
The integral of (1-sqrt(x+1))/(1+sqrt(x+1)) is -x+4sqrt(x+1)-4ln|sqrt(x+1)+1|+C