分母に(1+x²)²を含む積分
分母に(1+x²)²を含む積分
(1)
\[ \int\frac{1}{\left(1+x^{2}\right)^{2}}dx=\frac{1}{2}\tan^{\bullet}x+\frac{x}{2\left(1+x^{2}\right)}+C \](2)
\[ \int\frac{x}{\left(1+x^{2}\right)^{2}}dx=-\frac{1}{2\left(1+x^{2}\right)}+C \](3)
\[ \int\frac{x^{2}}{\left(1+x^{2}\right)^{2}}dx=\frac{1}{2}\tan^{\bullet}x-\frac{x}{2\left(1+x^{2}\right)}+C \](4)
\[ \int\frac{x^{3}}{\left(1+x^{2}\right)^{2}}dx=\frac{1}{2}\log\left(1+x^{2}\right)+\frac{1}{2\left(1+x^{2}\right)}+C \](5)
\[ \int\frac{x^{4}}{\left(1+x^{2}\right)^{2}}dx=x-\frac{3}{2}\tan^{\bullet}x+\frac{x}{2\left(1+x^{2}\right)}+C \](1)
\begin{align*} \int\frac{1}{\left(1+x^{2}\right)^{2}}dx & =\int\left(\frac{1+x^{2}}{\left(1+x^{2}\right)^{2}}-\frac{x^{2}}{\left(1+x^{2}\right)^{2}}\right)dx\\ & =\int\left(\frac{1}{1+x^{2}}+\frac{x}{2}\frac{-2x}{\left(1+x^{2}\right)^{2}}\right)dx\\ & =\tan^{\bullet}x+C+\frac{x}{2}\frac{1}{1+x^{2}}-\frac{1}{2}\int\left(\frac{1}{1+x^{2}}\right)dx\\ & =\tan^{\bullet}x+C+\frac{x}{2\left(1+x^{2}\right)}-\frac{1}{2}\tan^{\bullet}x\\ & =\frac{1}{2}\tan^{\bullet}x+\frac{x}{2\left(1+x^{2}\right)}+C \end{align*}(2)
\begin{align*} \int\frac{x}{\left(1+x^{2}\right)^{2}}dx & =\int\frac{1}{2\left(1+x^{2}\right)^{2}}d\left(1+x^{2}\right)\\ & =-\frac{1}{2\left(1+x^{2}\right)}+C \end{align*}(3)
\begin{align*} \int\frac{x^{2}}{\left(1+x^{2}\right)^{2}}dx & =\int\left(\frac{1+x^{2}}{\left(1+x^{2}\right)^{2}}-\frac{1}{\left(1+x^{2}\right)^{2}}\right)dx\\ & =\int\left(\frac{1}{1+x^{2}}-\frac{1}{\left(1+x^{2}\right)^{2}}\right)dx\\ & =\tan^{\bullet}x-\left(\frac{1}{2}\tan^{\bullet}x+\frac{x}{2\left(1+x^{2}\right)}\right)+C\\ & =\frac{1}{2}\tan^{\bullet}x-\frac{x}{2\left(1+x^{2}\right)}+C \end{align*}(4)
\begin{align*} \int\frac{x^{3}}{\left(1+x^{2}\right)^{2}}dx & =\int\left(\frac{x\left(x^{2}+1\right)}{\left(1+x^{2}\right)^{2}}-\frac{x}{\left(1+x^{2}\right)^{2}}\right)dx\\ & =\int\left(\frac{x}{1+x^{2}}-\frac{x}{\left(1+x^{2}\right)^{2}}\right)dx\\ & =\int\frac{1}{2\left(1+x^{2}\right)}d\left(1+x^{2}\right)-\int\frac{x}{\left(1+x^{2}\right)^{2}}dx\\ & =\frac{1}{2}\log\left(1+x^{2}\right)-\left(-\frac{1}{2\left(1+x^{2}\right)}\right)+C\\ & =\frac{1}{2}\log\left(1+x^{2}\right)+\frac{1}{2\left(1+x^{2}\right)}+C \end{align*}(5)
\begin{align*} \int\frac{x^{4}}{\left(1+x^{2}\right)^{2}}dx & =\int\left(\frac{\left(1+x^{2}\right)^{2}}{\left(1+x^{2}\right)^{2}}-\frac{2x^{2}+1}{\left(1+x^{2}\right)^{2}}\right)dx\\ & =\int\left(1-2\frac{x^{2}+1}{\left(1+x^{2}\right)^{2}}+\frac{1}{\left(1+x^{2}\right)^{2}}\right)dx\\ & =\int\left(1-2\frac{1}{\left(1+x^{2}\right)}+\frac{1}{\left(1+x^{2}\right)^{2}}\right)dx\\ & =x-2\tan^{\bullet}x+\frac{1}{2}\tan^{\bullet}x+\frac{x}{2\left(1+x^{2}\right)}+C\\ & =x-\frac{3}{2}\tan^{\bullet}x+\frac{x}{2\left(1+x^{2}\right)}+C \end{align*}ページ情報
| タイトル | 分母に(1+x²)²を含む積分 |
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\[
\int_{-\infty}^{\infty}\Gamma\left(1-ix\right)\Gamma\left(1+ix\right)dx=?
\]
気付かないと解けないかも
\[
\int_{0}^{\infty}\frac{1}{\left(1+x\right)\left(a^{2}+\log^{2}x\right)}dx=?
\]
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\[
\int\log^{\frac{1}{2}}xdx=x\log^{\frac{1}{2}}x-\frac{\sqrt{\pi}}{2}erfi\left(\log^{\frac{1}{2}}x\right)+C
\]
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\[
\int_{0}^{\infty}e^{-x}\log^{2}\left(x\right)dx=?
\]