x-x-2-4x-3-dx-

Question Number 171971 by ilhamQ last updated on 22/Jun/22

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

Answered by cortano1 last updated on 22/Jun/22

\frac{x}{(x + 1) (x + 3)} = \frac{a}{x + 1} + \frac{b}{x + 3}

a = \frac{- 1}{- 1 + 3} = - \frac{1}{2}

b = \frac{- 3}{- 3 + 1} = \frac{3}{2}

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

Answered by Mathspace last updated on 22/Jun/22

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

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

x^{2} + 4 x + 3 = 0 \to Δ^{^{'}} = 2^{2} - 3 = 1 \Rightarrow

x_{1} = \frac{- 2 + 1}{1} = - 1 a n d x_{2} = \frac{- 2 - 1}{1} = - 3 \Rightarrow

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

= \frac{1}{2} (\frac{1}{x + 1} - \frac{1}{x + 3}) \Rightarrow

\int \frac{d x}{x^{2} + 4 x + 3} = \frac{1}{2} l n ∣ \frac{x + 1}{x + 3} ∣ + c_{1}

\int \frac{2 x + 4}{x^{2} + 4 x + 3} d x = l n ∣ x^{2} + 4 x + 3 ∣ + c_{2}

\Rightarrow I = \frac{1}{2} l n ∣ x^{2} + 4 x + 3 ∣ - l n ∣ \frac{x + 1}{x + 3} ∣ + C