What is the integral of 1/(x+2sqrt(x)+3) ?

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

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integral of 1/(x+2sqrt(x)+3)

\int \frac{1}{x + 2 x + 3} d x

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

Solution steps

\int \frac{1}{x + 2 x + 3} d x

Apply u-substitution

: \int \frac{2 u}{u ^{2} + 2 u + 3} d u

= \int \frac{2 u}{u ^{2} + 2 u + 3} 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 u + 3} d u

Complete the square

u^{2} + 2 u + 3 : (u + 1)^{2} + 2

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

Apply u-substitution

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

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

Expand

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

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

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

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

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

Substitute back

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

Simplify

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

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

Add a constant to the solution

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

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