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 |
URL | https://www.nomuramath.com/cw47h8wx/ |
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tanの立方根の積分
\[
\int\sqrt[3]{\tan x}dx=\frac{1}{4}\log\left(\tan^{\frac{4}{3}}x-\tan^{\frac{2}{3}}x+1\right)+\frac{\sqrt{3}}{2}\tan^{\bullet}\left(\frac{2\tan^{\frac{2}{3}}x-1}{\sqrt{3}}\right)-\frac{1}{2}\log\left(\tan^{\frac{2}{3}}x+1\right)+C
\]
気付かないと解けないかも
\[
\int_{0}^{\infty}\frac{1}{\left(1+x\right)\left(a^{2}+\log^{2}x\right)}dx=?
\]
分母に正接がある関数の定積分
\[
\int_{0}^{\frac{\pi}{2}}\frac{x}{\tan x}dx=?
\]
分母に1乗と2乗ルートの積分
\[
\int\frac{1}{\left(z\pm1\right)\sqrt{z^{2}-1}}dz=\frac{\sqrt{z^{2}-1}}{\pm z+1}+C
\]