Find \(\int \frac{(x+3)}{2x^2+6x+7} dx\).

We can express
x+3=A \(\frac{d}{dx}\) (2x²+6x+7)+B
x+3=A(4x+6)+B
x+3=4Ax+(6A+B)
Comparing the coefficients, we get
4A=1 ?A=1/4
6A+B=3 ?B=3/2
\(\int \frac{x+3}{2x^2+6x+7} dx=\frac{1}{4} \int \frac{4x+6}{2x^2+6x+7} dx+\frac{3}{2} \int \frac{1}{2x^2+6x+7} dx\)
Let 2x²+6x+7=t
(4x+6)dx=dt
\(\frac{1}{4} \int \frac{4x+6}{2x^2+6x+7} dx=\frac{1}{4} \int \frac{dt}{t}=\frac{1}{4} log?t\)
Replacing t with (2x²+6x+7)
\(\frac{1}{4} \int \frac{4x+6}{2x^2+6x+7} dx=\frac{1}{4} log?(2x^2+6x+7)\)
\(\frac{3}{2} \int \frac{1}{2x^2+6x+7} dx=\frac{3}{2} \int \frac{1}{2(x^2+3x+\frac{7}{2})} dx=\frac{3}{4} \int \frac{1}{(x+\frac{3}{2})^2+2} dx\)
=\(\frac{3}{4} \left (\frac{1}{\sqrt{2}} tan^{-1}?\frac{2x+3}{2\sqrt{2}} \right )\)
?\(\int \frac{x+3}{2x^2+6x+7} dx=\frac{1}{4} log?(2x^2+6x+7) + \frac{3}{4} \left (\frac{1}{\sqrt{2}} tan^{-1}?\frac{2x+3}{2?2} \right )+C\)