What is the integral of (2x)/(x^2+4x+3) ?

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

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integral of (2x)/(x^2+4x+3)

\int \frac{2 x}{x ^{2} + 4 x + 3} d x

ln x^{2} + 4 x + 3 + 2 ln ∣ x + 3 ∣ - 2 ln ∣ x + 1 ∣ + C

Solution steps

\int \frac{2 x}{x ^{2} + 4 x + 3} 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} + 4 x + 3} d x

Complete the square

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

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

Apply u-substitution

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

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

Expand

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

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

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

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

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

Substitute back

u = x + 2

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

Simplify

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

= ln x^{2} + 4 x + 3 + 2 ln ∣ x + 3 ∣ - 2 ln ∣ x + 1 ∣

Add a constant to the solution

= ln x^{2} + 4 x + 3 + 2 ln ∣ x + 3 ∣ - 2 ln ∣ x + 1 ∣ + C

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

\int 2 - x - x^{2} d x

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

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

\int 15 e^{3 x} d x

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