What is the integral of 1/((1+sqrt(x))^2) ?

The integral of 1/((1+sqrt(x))^2) is 2(ln|1+sqrt(x)|+1/(1+sqrt(x)))+C

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integral of 1/((1+sqrt(x))^2)

\int \frac{1}{( 1 + x ) ^{2}} d x

2 (ln 1 + x + \frac{1}{1 + x}) + C

Solution steps

\int \frac{1}{( 1 + x ) ^{2}} d x

Apply u-substitution

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

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

Apply u-substitution

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

= 2 \cdot \int \frac{v - 1}{v ^{2}} d v

Expand

\frac{v - 1}{v ^{2}} : \frac{1}{v} - \frac{1}{v ^{2}}

= 2 \cdot \int \frac{1}{v} - \frac{1}{v ^{2}} d v

Apply the Sum Rule:

\int f (x) \pm g (x) d x = \int f (x) d x \pm \int g (x) d x

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

\int \frac{1}{v} d v = ln ∣ v ∣

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

= 2 (ln ∣ v ∣ - (- \frac{1}{v}))

Substitute back

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

Simplify

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

Add a constant to the solution

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

y = \frac{x - x}{x ^{2}}

ln (1 + 3 x)

\frac{d y}{d x} + 2 x y = y + 6 x - 3

\frac{d}{d x} (\frac{2}{3 x})

What is the integral of 1/((1+sqrt(x))^2) ?
The integral of 1/((1+sqrt(x))^2) is 2(ln|1+sqrt(x)|+1/(1+sqrt(x)))+C