詳解：分母不可分解，需將分母配方

\({x^2} + 2x + 2 = ({x^2} + 2x + 1) + 1 = {(x + 1)^2} + 1\)

\(\int {\frac{{4x + 7}}{{{x^2} + 2x + 2}}\;dx}  = \int {\frac{{4x + 7}}{{{{(x + 1)}^2} + 1}}} dx\)

強迫分子遷就分母改 \((x + 1)\) 表示

\(\int {\frac{{4x + 7}}{{{{(x + 1)}^2} + 1}}} dx = \int {\frac{{4(x + 1) + 3}}{{{{(x + 1)}^2} + 1}}} dx\)

分解成為兩個部分

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

前面部分可利用口訣 " 分母的微分 = 分子 " ， 則其積分為 ln (分母)

\(2\int {\frac{{2(x + 1)}}{{{{(x + 1)}^2} + 1}}} dx = 2\ln \left[ {{{(x + 1)}^2} + 1} \right] + {C_1} = 2\ln ({x^2} + 2x + 2) + {C_1}\)

後面部分使用基本積分

\(\int {\frac{1}{{1 + {x^2}}}dx}  = Ta{n^{ - 1}}x + C\)

\(\int {\frac{3}{{{{(x + 1)}^2} + 1}}} dx = 3Ta{n^{ - 1}}(x + 1) + {C_2}\)

合併兩部分得到答案

\(\int {\frac{{4x + 7}}{{{x^2} + 2x + 2}}\;dx}  = 2\ln ({x^2} + 2x + 2) + 3Ta{n^{ - 1}}(x + 1) + C\)