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

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

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

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

\frac{1}{2} ln 2 x^{2} + 22 x + 2 + arctan (2 x + 1) + C

Solution steps

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

Complete the square

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

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

Apply u-substitution

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

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

Take the constant out:

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

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

Expand

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

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

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

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

Substitute back

u = x + \frac{1}{2}

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

Simplify

2 (\frac{1}{2 2} ln 2 (x + \frac{1}{2})^{2} + 1 + \frac{1}{2} arctan (2 (x + \frac{1}{2}))) : \frac{1}{2} ln 2 x^{2} + 22 x + 2 + arctan (2 x + 1)

= \frac{1}{2} ln 2 x^{2} + 22 x + 2 + arctan (2 x + 1)

Add a constant to the solution

= \frac{1}{2} ln 2 x^{2} + 22 x + 2 + arctan (2 x + 1) + C

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