What is the integral of 1/(x^2+4x+2) ?

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

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integral of 1/(x^2+4x+2)

\int \frac{1}{x ^{2} + 4 x + 2} d x

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

Solution steps

\int \frac{1}{x ^{2} + 4 x + 2} d x

Complete the square

x^{2} + 4 x + 2 : (x + 2)^{2} - 2

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

Apply u-substitution

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

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

Apply Integral Substitution

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

Take the constant out:

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

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

Factor

v^{2} - 1 : - (- v^{2} + 1)

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

Take the constant out:

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

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

Use the common integral:

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

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

Substitute back

= \frac{1}{2} - \frac{ln \frac{x + 2}{2} + 1}{2} - \frac{ln \frac{x + 2}{2} - 1}{2}

Simplify

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

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

Add a constant to the solution

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

x^{3} y^{^{'}} + 2 x^{2} y = 4 cos (4 x) y^{- \frac{1}{2}}

f (x) = lo g_{a} (x)

\frac{\partial}{\partial x} (x^{\frac{1}{2}} + y)

y = 4 e^{x}

\frac{\partial}{\partial t} (t^{2})

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