What is the integral of (x+5)/(x^2+4x+3) ?

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

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integral of (x+5)/(x^2+4x+3)

\int \frac{x + 5}{x ^{2} + 4 x + 3} d x

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

Solution steps

\int \frac{x + 5}{x ^{2} + 4 x + 3} d x

Apply u-substitution

: \int \frac{u}{u ^{2} - 6 u + 8} d u

= \int \frac{u}{u ^{2} - 6 u + 8} d u

Complete the square

u^{2} - 6 u + 8 : (u - 3)^{2} - 1

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

Apply u-substitution

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

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

Expand

\frac{v + 3}{v ^{2} - 1} : \frac{v}{v ^{2} - 1} + \frac{3}{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

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

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

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

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

Substitute back

= \frac{1}{2} ln (x + 5 - 3)^{2} - 1 - 3 (\frac{1}{2} ln ∣ x + 5 - 3 + 1 ∣ - \frac{1}{2} ln ∣ x + 5 - 3 - 1 ∣)

Simplify

\frac{1}{2} ln (x + 5 - 3)^{2} - 1 - 3 (\frac{1}{2} ln ∣ x + 5 - 3 + 1 ∣ - \frac{1}{2} ln ∣ x + 5 - 3 - 1 ∣) : \frac{1}{2} ln x^{2} + 4 x + 3 - 3 (\frac{1}{2} ln ∣ x + 3 ∣ - \frac{1}{2} ln ∣ x + 1 ∣)

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

Add a constant to the solution

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

x \to - 8 lim (x + 6)

\int \frac{7 ln ( x + 8 )}{x ^{2}} d x

\frac{d y}{d x} - 2 x y = 10, y (0) = - 8

\int_{2}^{4} 4 (3 x - 1) d x

y^{^{''}} - 100 y = 0

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