テキスト（簡単な置換積分）

\(u = x + 1\) とおくと，\(\displaystyle \frac{du}{dx} = 1\)

与式 \(\begin{array}[t]{l} \displaystyle = \int (u - 1)u^4\cdot\frac{1}{1}\,du \\ \displaystyle = \int(u^5 - u^4)\,du \\ \displaystyle = \frac{1}{6}u^6 - \frac{1}{5}u^5 + C \\ \displaystyle = u^5\left(\frac{1}{6}u - \frac{1}{5}\right) + C \\ \displaystyle = \frac{1}{30}u^5(5u - 6) + C \\ \displaystyle = \frac{1}{30}(x + 1)\big\{5(x + 1) - 6\big\} \\ \displaystyle = \frac{1}{30}(5x - 1)(x + 1)^5 + C \end{array}\)

\(u = x - 1\) とおくと，\(\displaystyle \frac{du}{dx} = 1\)

与式 \(\begin{array}[t]{l} \displaystyle = \int \big\{(u + 1) + 1\big\}u^6\,du \\ \displaystyle = \int (u + 2)u^6\,du \\ \displaystyle = \int \left(u^7 + 2u^6\right)\,du \\ \displaystyle = \frac{1}{8}u^8 + \frac{2}{7}u^7 + C \\ \displaystyle = \frac{1}{56}u^7(7u + 16) + C \\ \displaystyle = \frac{1}{56}(x - 1)^7\big\{7(x - 1) + 16\big\} + C \\ \displaystyle = \frac{1}{56}(7x + 9)(x - 1)^7 + C \end{array}\)

\(u = x^2 + 1\) とおくと，\(\displaystyle \frac{du}{dx} = 2x\)

与式 \(\begin{array}[t]{l} \displaystyle = \int \cancel{x}\cdot u^3\cdot\frac{1}{2\cancel{x}}\,du \\ \displaystyle = \int \frac{1}{2}u^3\,du \\ \displaystyle = \frac{1}{8}u^4 + C \\ \displaystyle = \frac{1}{8}(x^2 + 1)^4 + C \end{array}\)

\(u = x^2 + 3\) とおくと，\(\displaystyle \frac{du}{dx} = 2x\)

与式 \(\begin{array}[t]{l} \displaystyle = \int \cancel{x}\sqrt{u}\cdot\frac{1}{2\cancel{x}}\,du \\ \displaystyle = \int \frac{1}{2}u^{\frac{1}{2}}\,du \\ \displaystyle = \frac{1}{2}\cdot\frac{1}{\frac{1}{2} + 1}u^{\frac{1}{2} + 1} + C \\ \displaystyle = \frac{1}{3}u\sqrt{u} + C \\ \displaystyle = \frac{1}{3}(x^2 + 3)\sqrt{x^2 + 3} + C \end{array}\)

\(u = x^2 + x + 1\) とおくと，\(\displaystyle \frac{du}{dx} = 2x + 1\)

与式 \(\begin{array}[t]{l} \displaystyle = \int \left(\cancel{2x + 1}\right)u^4\cdot\frac{1}{\cancel{2x + 1}}\,du \\ \displaystyle = \int u^4\,du \\ \displaystyle = \frac{1}{5}u^5 + C \\ \displaystyle = \frac{1}{5}(x^2 + x + 1)^5 + C \end{array}\)

\(u = x - 1\) とおくと，\(\displaystyle \frac{du}{dx} = 1\)

\(\begin{array}{c|ccc} x & 1 & \to & 2 \\ \hline u & 0 & \to & 1 \end{array}\)

与式 \(\begin{array}[t]{l} \displaystyle = \int_0^1 u^5 \,du = \left[\frac{1}{6}u^6\right]_0^1 = \frac{1}{6} \end{array}\)

\(\displaystyle u = \frac{2}{3}x + \frac{1}{3}\) とおくと，\(\displaystyle \frac{du}{dx} = \frac{2}{3}\)

\(\begin{array}{c|ccc} x & -2 & \to & 1 \\ \hline u & -1 & \to & 1 \end{array}\)

与式 \(\begin{array}[t]{l} \displaystyle = \int_{-1}^1 u^4\cdot\frac{3}{2}\,du = \left[\frac{3}{2}\cdot\frac{1}{5}u^5\right]_{-1}^1 = \frac{3}{10}\big\{1 - (-1)\big) = \frac{3}{5} \end{array}\)

\(\displaystyle u = \frac{1}{2}x + \frac{3}{2}\) とおくと，\(\displaystyle \frac{du}{dx} = \frac{1}{2}\)

\(\begin{array}{c|ccc} x & -1 & \to & 5 \\ \hline u & 1 & \to & 4 \end{array}\)

与式 \(\begin{array}[t]{l} \displaystyle = \int_1^4 u^{\frac{1}{2}}\cdot 2\,du \displaystyle = \left[2\cdot\frac{2}{3}u^{\frac{3}{2}}\right]_1^4 \displaystyle = \frac{4}{3}(8 - 1) \displaystyle = \frac{28}{3} \end{array}\)

\(u = x - 1\) とおくと，\(\displaystyle \frac{du}{dx} = 1\)

\(\begin{array}{c|ccc} x & 1 & \to & 2 \\ \hline u & 0 & \to & 1 \end{array}\)

与式 \(\begin{array}[t]{l} \displaystyle = \int_0^1(u + 1)u^4\,du \\ \displaystyle = \int_0^1(u^5 + u^4)\,du \\ \displaystyle = \left[\frac{1}{6}u^6 + \frac{1}{5}u^5\right]_0^1 \\ \displaystyle = \frac{1}{6} + \frac{1}{5} \\ \displaystyle = \frac{11}{30} \end{array}\)

\(u = x^2 + 1\) とおくと，\(\displaystyle \frac{du}{dx} = 2x\)

\(\begin{array}{c|ccc} x & 0 & \to & \sqrt{3} \\ \hline u & 1 & \to & 4 \end{array}\)

与式 \(\begin{array}[t]{l} \displaystyle = \int_1^4 \frac{\cancel{2x}}{\sqrt{u}}\cdot\frac{1}{\cancel{2x}}\,du \\ \displaystyle = \int_1^4 u^{-\frac{1}{2}}\,du \\ \displaystyle = \Big[2u^{\frac{1}{2}}\Big]_1^4 \\ \displaystyle = 2(2 - 1) \\ \displaystyle = 2 \end{array}\)