What is the integral of (x+4)/(x^2+5x+6) ?

The integral of (x+4)/(x^2+5x+6) is 2(1/4 ln|4x^2+20x+24|-3/4 (ln|2x+6|-ln|2x+4|))+C

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integral of (x+4)/(x^2+5x+6)

\int \frac{x + 4}{x ^{2} + 5 x + 6} d x

2 (\frac{1}{4} ln 4 x^{2} + 20 x + 24 - \frac{3}{4} (ln ∣ 2 x + 6 ∣ - ln ∣ 2 x + 4 ∣)) + C

Solution steps

\int \frac{x + 4}{x ^{2} + 5 x + 6} d x

Apply u-substitution

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

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

Complete the square

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

= \int \frac{u}{( u - \frac{3}{2} ) ^{2} - \frac{1}{4}} d u

Apply u-substitution

: \int \frac{2 ( 2 v + 3 )}{4 v ^{2} - 1} d v

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

Expand

\frac{2 v + 3}{4 v ^{2} - 1} : \frac{2 v}{4 v ^{2} - 1} + \frac{3}{4 v ^{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 v}{4 v ^{2} - 1} d v + \int \frac{3}{4 v ^{2} - 1} d v)

\int \frac{2 v}{4 v ^{2} - 1} d v = \frac{1}{4} ln 4 v^{2} - 1

\int \frac{3}{4 v ^{2} - 1} d v = - \frac{3}{4} (ln ∣ 2 v + 1 ∣ - ln ∣ 2 v - 1 ∣)

= 2 (\frac{1}{4} ln 4 v^{2} - 1 - \frac{3}{4} (ln ∣ 2 v + 1 ∣ - ln ∣ 2 v - 1 ∣))

Substitute back

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

Simplify

2 (\frac{1}{4} ln 4 (x + 4 - \frac{3}{2})^{2} - 1 - \frac{3}{4} (ln 2 (x + 4 - \frac{3}{2}) + 1 - ln 2 (x + 4 - \frac{3}{2}) - 1)) : 2 (\frac{1}{4} ln 4 x^{2} + 20 x + 24 - \frac{3}{4} (ln ∣ 2 x + 6 ∣ - ln ∣ 2 x + 4 ∣))

= 2 (\frac{1}{4} ln 4 x^{2} + 20 x + 24 - \frac{3}{4} (ln ∣ 2 x + 6 ∣ - ln ∣ 2 x + 4 ∣))

Add a constant to the solution

= 2 (\frac{1}{4} ln 4 x^{2} + 20 x + 24 - \frac{3}{4} (ln ∣ 2 x + 6 ∣ - ln ∣ 2 x + 4 ∣)) + C

\int \frac{4 s}{( 1 - 2 s ^{2} ) ^{2}} d s

\int e^{t a n (x^{2})} \cdot x \cdot sec^{2} (x^{2}) d x

\int (y^{2})^{2} d y

\int \frac{e ^{4 x}}{e ^{4 x} + 7} d x

What is the integral of (x+4)/(x^2+5x+6) ?
The integral of (x+4)/(x^2+5x+6) is 2(1/4 ln|4x^2+20x+24|-3/4 (ln|2x+6|-ln|2x+4|))+C