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

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

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

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

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

Solution steps

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

Take the constant out:

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

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

Complete the square

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

= 2 \cdot \int \frac{x}{( 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

= 2 \cdot \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

= 22 \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

= 22 (\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)

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

Substitute back

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

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

Simplify

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

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

Add a constant to the solution

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

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(2, 3), (- 2, - 3)

\frac{d}{d x} (\frac{4}{2 x + 1})

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