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

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

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integral of 2/((x^2+3x+2))

integral

\frac{2}{( x ^{2} + 3 x + 2 )}

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

Solution steps

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

Take the constant out:

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

= 2 \cdot \int \frac{1}{x ^{2} + 3 x + 2} d x

Complete the square

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

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

Apply u-substitution

: \int \frac{4}{4 u ^{2} - 1} d u

= 2 \cdot \int \frac{4}{4 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 4 \cdot \int \frac{1}{4 u ^{2} - 1} d u

Factor

4 u^{2} - 1 : - (- 4 u^{2} + 1)

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

Apply Integral Substitution

= 2 \cdot 4 (- \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

= 2 \cdot 4 (- \frac{1}{2} \cdot \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}

= 2 \cdot 4 (- \frac{1}{2} (\frac{ln ∣ v + 1 ∣}{2} - \frac{ln ∣ v - 1 ∣}{2}))

Substitute back

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

Simplify

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

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

Add a constant to the solution

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

\frac{1}{( 2 x ^{2} + 6 x - 8 )}

x (x - 1) (x + 1)

\frac{1}{( x ^{2} + 0.75 )}

f (x) = (1 + cos (6 x))^{\frac{1}{2}}

2 (sin (x))^{3} + sin (x) + tan^{5} (x)

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