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

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

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

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

ln ∣ x + 1 ∣ + C

Solution steps

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

Apply u-substitution

: \int \frac{1}{u - 3} d u

= \int \frac{1}{u - 3} d u

Apply u-substitution

: \int \frac{1}{v} d v

= \int \frac{1}{v} d v

Use the common integral:

\int \frac{1}{v} d v = ln (∣ v ∣)

= ln ∣ v ∣

Substitute back

= ln ∣ x + 4 - 3 ∣

Simplify

= ln ∣ x + 1 ∣

Add a constant to the solution

= ln ∣ x + 1 ∣ + C

x \to 3 + lim (ln (x^{2} - 9))

(- \frac{1}{( 1 + x ) ^{2}})^{^{'}}

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

\frac{d}{d x} (arccos (3 x))

t \frac{d y}{d t} + 21 y = \frac{1}{t}

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