What is the integral of (x+1)/(x^2+4x+3) ?

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

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integral of (x+1)/(x^2+4x+3)

\int \frac{x + 1}{x ^{2} + 4 x + 3} d x

ln ∣ x + 3 ∣ + C

Solution steps

\int \frac{x + 1}{x ^{2} + 4 x + 3} d x

Complete the square

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

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

Apply u-substitution

: \int \frac{1}{u + 1} d u

= \int \frac{1}{u + 1} 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 + 2 + 1 ∣

Simplify

= ln ∣ x + 3 ∣

Add a constant to the solution

= ln ∣ x + 3 ∣ + C

\int_{6}^{8} \frac{30}{( x - 6 ) ^{3}} d x

\int_{0}^{x} (x + 2 y) d y

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

\int x sin (\frac{1}{3}) x d x

t y^{^{'}} + 4 y = t^{2} - t + 5

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