積分

計算と説明をお願いします

$$ \begin{aligned} \frac{7x^2 + 3x + 16}{(x^2+3)(2x+1)} = \frac{ax+b}{x^2+3} + \frac{c}{2x+1} \end{aligned} $$ とおき，右辺を整理すると， $$ \begin{aligned} \frac{7x^2 + 3x + 16}{(x^2+3)(2x+1)} =　\frac{(2a+c)x^2 + (a + 2b)x +(b + 3c)}{(x^2+3)(2x+1)} \end{aligned} $$ となる．両辺の係数を比較して $$ \begin{align} \left\{ \begin{array}{ll} 7 &= 2a+c\\ 3 &= a+2b \\ 16 &= b+3c \end{array} \right. \end{align} $$ これを解くと， $$ \begin{aligned} (a, b, c) = (1, 1, 5) \end{aligned} $$ となる．よって， $$ \begin{aligned} \int_{1}^{3} \frac{7x^2 + 3x + 16}{(x^2+3)(2x+1)} dx &= \int_{1}^{3}( \frac{x+1}{x^2+3} + \frac{5}{2x+1} )dx\\ &= \int_{1}^{3}\frac{x}{x^2+3} dx + \int_{1}^{3}\frac{1}{x^2+3} dx + \int_{1}^{3}\frac{5}{2x+1} dx \end{aligned} $$ ここで， $$ \begin{aligned} \int_{1}^{3}\frac{x}{x^2+3} dx &= \int_{1}^{3} \frac{1}{2} \frac{(x^2+3)'}{x^2+3} dx \\ &=\frac{1}{2} [\log{(x^2+3)}]_{1}^{3}\\ &= \frac{1}{2} \log 3 \end{aligned} $$ $$ \begin{aligned} \int_{1}^{3} \frac{5}{2x+1} dx &= \int_{1}^{3} \frac{5}{2} \frac{(2x+1)'}{2x+1} dx \\ &=\frac{5}{2} [\log{(2x+1)}]_{1}^{3}\\ &= \frac{5}{2} (\log 7 - \log 3) \end{aligned} $$ $$ \begin{aligned} \int_{1}^{3}\frac{1}{x^2+3} dx &= \int_{\frac{\pi}{6}}^{\frac{\pi}{3}} \frac{1}{3 ( \tan ^2 \theta+1 )} dx 　(x = \sqrt{3} \tan \theta で置換) \\ &= \frac{1}{3} \int_{\frac{\pi}{6}}^{\frac{\pi}{3}} \cos^2 \theta dx \\ &= \frac{1}{3} \int_{\frac{\pi}{6}}^{\frac{\pi}{3}} \frac{\cos 2\theta + 1}{2} \theta dx \\ ・\\ ・\\ ・\\ \end{aligned} $$ という感じで計算できると思います．