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

The integral of 1/(x+1+sqrt(x)) is ln|4x+4sqrt(x)+4|-2/(sqrt(3))arctan((2sqrt(x)+1)/(sqrt(3)))+C

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

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

ln 4 x + 4 x + 4 - \frac{2}{3} arctan (\frac{2 x + 1}{3}) + C

Solution steps

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

Apply u-substitution

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

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

Complete the square

u^{2} + 1 + u : (u + \frac{1}{2})^{2} + \frac{3}{4}

= 2 \cdot \int \frac{u}{( u + \frac{1}{2} ) ^{2} + \frac{3}{4}} d u

Apply u-substitution

: \int \frac{2 ( 2 v - 1 )}{4 v ^{2} + 3} d v

= 2 \cdot \int \frac{2 ( 2 v - 1 )}{4 v ^{2} + 3} d v

Take the constant out:

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

= 2 \cdot 2 \cdot \int \frac{2 v - 1}{4 v ^{2} + 3} d v

Expand

\frac{2 v - 1}{4 v ^{2} + 3} : \frac{2 v}{4 v ^{2} + 3} - \frac{1}{4 v ^{2} + 3}

Apply the Sum Rule:

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

= 2 \cdot 2 (\int \frac{2 v}{4 v ^{2} + 3} d v - \int \frac{1}{4 v ^{2} + 3} d v)

\int \frac{2 v}{4 v ^{2} + 3} d v = \frac{1}{4} ln 4 v^{2} + 3

\int \frac{1}{4 v ^{2} + 3} d v = \frac{1}{2 3} arctan (\frac{2}{3} v)

= 2 \cdot 2 (\frac{1}{4} ln 4 v^{2} + 3 - \frac{1}{2 3} arctan (\frac{2}{3} v))

Substitute back

= 2 \cdot 2 (\frac{1}{4} ln 4 (x + \frac{1}{2})^{2} + 3 - \frac{1}{2 3} arctan (\frac{2}{3} (x + \frac{1}{2})))

Simplify

2 \cdot 2 (\frac{1}{4} ln 4 (x + \frac{1}{2})^{2} + 3 - \frac{1}{2 3} arctan (\frac{2}{3} (x + \frac{1}{2}))) : ln 4 x + 4 x + 4 - \frac{2}{3} arctan (\frac{2 x + 1}{3})

= ln 4 x + 4 x + 4 - \frac{2}{3} arctan (\frac{2 x + 1}{3})

Add a constant to the solution

= ln 4 x + 4 x + 4 - \frac{2}{3} arctan (\frac{2 x + 1}{3}) + C

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What is the integral of 1/(x+1+sqrt(x)) ?
The integral of 1/(x+1+sqrt(x)) is ln|4x+4sqrt(x)+4|-2/(sqrt(3))arctan((2sqrt(x)+1)/(sqrt(3)))+C