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

$$\int \frac{x}{x^2+4x+3}dx=...$$

Answered by cortano1 last updated on 22/Jun/22

$$\frac{x}{(x+1)(x+3)}=\frac{a}{x+1}+\frac{b}{x+3}$$$$a=\frac{-1}{-1+3}=-\frac{1}{2}$$$$b=\frac{-3}{-3+1}=\frac{3}{2}$$$$\int \frac{x}{x^2+4x+3}dx=-\frac{1}{2}\ln \mid x+1\mid +\frac{3}{2}\ln \mid x+3\mid +c$$

Answered by Mathspace last updated on 22/Jun/22

$$I=\frac{1}{2}\int \frac{2x+4-4}{x^2+4x+3}dx$$$$=\frac{1}{2}\int \frac{2x+4}{x^2+4x+3}dx-2\int \frac{dx}{x^2+4x+3}$$$$x^2+4x+3=0\to {\mathrm{\Delta }}^{{}^{\prime }}=2^2-3=1\Rightarrow$$$$x_1=\frac{-2+1}{1}=-1and{x}_2=\frac{-2-1}{1}=-3\Rightarrow$$$$\frac{1}{x^2+4x+3}=\frac{1}{(x+1)(x+3)}$$$$=\frac{1}{2}(\frac{1}{x+1}-\frac{1}{x+3})\Rightarrow$$$$\int \frac{dx}{x^2+4x+3}=\frac{1}{2}ln\mid \frac{x+1}{x+3}\mid +c_1$$$$\int \frac{2x+4}{x^2+4x+3}dx=ln\mid x^2+4x+3\mid +c_2$$$$\Rightarrow I=\frac{1}{2}ln\mid x^2+4x+3\mid -ln\mid \frac{x+1}{x+3}\mid +C$$

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