What is the integral of (2x+1)/(x^2+5x+6) ?

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

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integral of (2x+1)/(x^2+5x+6)

\int \frac{2 x + 1}{x ^{2} + 5 x + 6} d x

ln 4 x^{2} + 20 x + 24 + 4 ln ∣ 2 x + 6 ∣ - 4 ln ∣ 2 x + 4 ∣ + C

Solution steps

\int \frac{2 x + 1}{x ^{2} + 5 x + 6} d x

Complete the square

x^{2} + 5 x + 6 : (x + \frac{5}{2})^{2} - \frac{1}{4}

= \int \frac{2 x + 1}{( x + \frac{5}{2} ) ^{2} - \frac{1}{4}} d x

Apply u-substitution

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

= \int \frac{4 ( 2 u - 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

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

Expand

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

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

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

\int \frac{4}{4 u ^{2} - 1} d u = - ln ∣ 2 u + 1 ∣ + ln ∣ 2 u - 1 ∣

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

Substitute back

u = x + \frac{5}{2}

= 4 (\frac{1}{4} ln 4 (x + \frac{5}{2})^{2} - 1 - (- ln 2 (x + \frac{5}{2}) + 1 + ln 2 (x + \frac{5}{2}) - 1))

Simplify

4 (\frac{1}{4} ln 4 (x + \frac{5}{2})^{2} - 1 - (- ln 2 (x + \frac{5}{2}) + 1 + ln 2 (x + \frac{5}{2}) - 1)) : ln 4 x^{2} + 20 x + 24 + 4 ln ∣ 2 x + 6 ∣ - 4 ln ∣ 2 x + 4 ∣

= ln 4 x^{2} + 20 x + 24 + 4 ln ∣ 2 x + 6 ∣ - 4 ln ∣ 2 x + 4 ∣

Add a constant to the solution

= ln 4 x^{2} + 20 x + 24 + 4 ln ∣ 2 x + 6 ∣ - 4 ln ∣ 2 x + 4 ∣ + C

\frac{d}{d x} (2 x^{2} + 5)

\int 4 x^{2} e^{3 x} d x

y^{^{''}} + y = 3 e^{- 4 x}

\int \frac{3 ln ^{2} ( x )}{x ^{2}} d x

\frac{d}{d x} (\frac{x sec ( - ( \frac{y}{x} ) + y )}{x})

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