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

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

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

\int \frac{1}{x + 2 + 2 x} d x

ln x + 2 x + 2 - 2 arctan (x + 1) + C

Solution steps

\int \frac{1}{x + 2 + 2 x} d x

Apply u-substitution

: \int \frac{2 u}{u ^{2} + 2 + 2 u} d u

= \int \frac{2 u}{u ^{2} + 2 + 2 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}{u ^{2} + 2 + 2 u} d u

Complete the square

u^{2} + 2 + 2 u : (u + 1)^{2} + 1

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

Apply u-substitution

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

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

Expand

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

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

\int \frac{v}{v ^{2} + 1} d v = \frac{1}{2} ln v^{2} + 1

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

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

Substitute back

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

Simplify

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

= ln x + 2 x + 2 - 2 arctan (x + 1)

Add a constant to the solution

= ln x + 2 x + 2 - 2 arctan (x + 1) + C

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