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

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

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

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

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

Solution steps

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

Complete the square

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

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

Apply u-substitution

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

= \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

= \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 ∣

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

Substitute back

u = x + 2

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

Simplify

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

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

Add a constant to the solution

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

\int \frac{1}{4 - y} d y

x \to 8 + lim (\frac{x - 8}{∣ x - 8 ∣})

y = 3 x^{2} - 2 x, at x = - 1

\frac{d}{d x} (x^{2} (x - 2)^{4})

\frac{\partial}{\partial x} (2 x + 3 y)

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