$$$\frac{1}{\left(a^{2} + x^{2}\right)^{\frac{3}{2}}}$$$ の $$$x$$$ に関する積分
関連する計算機: 定積分・広義積分計算機
入力内容
$$$\int \frac{1}{\left(a^{2} + x^{2}\right)^{\frac{3}{2}}}\, dx$$$ を求めよ。
解答
$$$x=\sinh{\left(u \right)} \left|{a}\right|$$$ とする。
すると $$$dx=\left(\sinh{\left(u \right)} \left|{a}\right|\right)^{\prime }du = \cosh{\left(u \right)} \left|{a}\right| du$$$ (手順は»で確認できます)。
また、$$$u=\operatorname{asinh}{\left(\frac{x}{\left|{a}\right|} \right)}$$$が成り立つ。
被積分関数は次のようになる
$$$\frac{1}{\left(a^{2} + x^{2}\right)^{\frac{3}{2}}} = \frac{1}{\left(a^{2} \sinh^{2}{\left( u \right)} + a^{2}\right)^{\frac{3}{2}}}$$$
恒等式 $$$\sinh^{2}{\left( u \right)} + 1 = \cosh^{2}{\left( u \right)}$$$ を用いよ:
$$$\frac{1}{\left(a^{2} \sinh^{2}{\left( u \right)} + a^{2}\right)^{\frac{3}{2}}}=\frac{\left|{a}\right|}{a^{4} \left(\sinh^{2}{\left( u \right)} + 1\right)^{\frac{3}{2}}}=\frac{\left|{a}\right|}{a^{4} \left(\cosh^{2}{\left( u \right)}\right)^{\frac{3}{2}}}$$$
$$$\frac{\left|{a}\right|}{a^{4} \left(\cosh^{2}{\left( u \right)}\right)^{\frac{3}{2}}} = \frac{\left|{a}\right|}{a^{4} \cosh^{3}{\left( u \right)}}$$$
したがって、
$${\color{red}{\int{\frac{1}{\left(a^{2} + x^{2}\right)^{\frac{3}{2}}} d x}}} = {\color{red}{\int{\frac{1}{a^{2} \cosh^{2}{\left(u \right)}} d u}}}$$
定数倍の法則 $$$\int c f{\left(u \right)}\, du = c \int f{\left(u \right)}\, du$$$ を、$$$c=\frac{1}{a^{2}}$$$ と $$$f{\left(u \right)} = \frac{1}{\cosh^{2}{\left(u \right)}}$$$ に対して適用する:
$${\color{red}{\int{\frac{1}{a^{2} \cosh^{2}{\left(u \right)}} d u}}} = {\color{red}{\frac{\int{\frac{1}{\cosh^{2}{\left(u \right)}} d u}}{a^{2}}}}$$
被積分関数を双曲線正割関数で表せ:
$$\frac{{\color{red}{\int{\frac{1}{\cosh^{2}{\left(u \right)}} d u}}}}{a^{2}} = \frac{{\color{red}{\int{\operatorname{sech}^{2}{\left(u \right)} d u}}}}{a^{2}}$$
$$$\operatorname{sech}^{2}{\left(u \right)}$$$ の不定積分は $$$\int{\operatorname{sech}^{2}{\left(u \right)} d u} = \tanh{\left(u \right)}$$$ です:
$$\frac{{\color{red}{\int{\operatorname{sech}^{2}{\left(u \right)} d u}}}}{a^{2}} = \frac{{\color{red}{\tanh{\left(u \right)}}}}{a^{2}}$$
次のことを思い出してください $$$u=\operatorname{asinh}{\left(\frac{x}{\left|{a}\right|} \right)}$$$:
$$\frac{\tanh{\left({\color{red}{u}} \right)}}{a^{2}} = \frac{\tanh{\left({\color{red}{\operatorname{asinh}{\left(\frac{x}{\left|{a}\right|} \right)}}} \right)}}{a^{2}}$$
したがって、
$$\int{\frac{1}{\left(a^{2} + x^{2}\right)^{\frac{3}{2}}} d x} = \frac{x}{a^{2} \sqrt{1 + \frac{x^{2}}{a^{2}}} \left|{a}\right|}$$
簡単化せよ:
$$\int{\frac{1}{\left(a^{2} + x^{2}\right)^{\frac{3}{2}}} d x} = \frac{x}{a^{2} \sqrt{a^{2} + x^{2}}}$$
積分定数を加える:
$$\int{\frac{1}{\left(a^{2} + x^{2}\right)^{\frac{3}{2}}} d x} = \frac{x}{a^{2} \sqrt{a^{2} + x^{2}}}+C$$
解答
$$$\int \frac{1}{\left(a^{2} + x^{2}\right)^{\frac{3}{2}}}\, dx = \frac{x}{a^{2} \sqrt{a^{2} + x^{2}}} + C$$$A