積分公式-03

分母に変数がある場合

　$ \Large \displaystyle \int \frac{1}{(1 + x)^2} \ dx $

$$
{ \Large \displaystyle
\begin{equation}
\begin{split}
\int \frac{1}{(1 + x)^2} \ dx \\
u &= 1+x -> du = dx \\
\int x^n \ dx &= \frac{x^{n+1}}{n+1} + C \\
\int \frac{1}{u^2} \ du &= \int u^{-2} \ du = - u^{-1} \\
\int \frac{1}{(1 + x)^2} \ dx &= -\frac{1}{1 + x} + C\\
\end{split}
\end{equation}
}
$$

$ \Large \displaystyle\int \frac{1}{1 + x^2} \ dx $

$$
{ \Large \displaystyle
\begin{equation}
\begin{split}
\int \frac{1}{1 + x^2} \ dx \\
x &= tan \ \theta \\
\left( \frac{f}{g} \right)' &= \frac{f'}{g} - \frac{fg'}{g^2} = \frac{f'g - fg'}{g^2}\\
\frac{du}{ d \theta} &= \frac{d}{ d \theta}tan \ \theta = \frac{d}{ d \theta} \frac{sin \ \theta}{cos \ \theta} \\
f &= sin \ \theta, \ g = cos \ \theta \\
f' &= cos \ \theta, \ g' = -sin \ \theta \\
&= \frac{cos \ \theta \cdot cos \ \theta - sin \ \theta (-sin \ \theta)}{cos^2 \ \theta} \\
&= \frac{1}{cos^2 \ \theta} \\
dx &= \frac{1}{cos^2 \ \theta} d \theta\\
1 + tan^2 \ \theta &= \frac{cos^2 \ \theta}{cos^2 \ \theta} + \frac{sin^2 \ \theta}{cos^2 \ \theta} = \frac{1}{cos^2 \ \theta} \\
1 + x^2 &= 1 + tan^2 \ \theta= \frac{1}{cos^2 \ \theta} \\
\int \frac{1}{1 + x^2} \ dx &= \int cos^2 \ \theta \frac{1}{cos^2 \ \theta} d \theta\\
&= \int d \theta = \theta +C \\
\int \frac{1}{1 + x^2} \ dx &= tan^{-1} x +C\\
\end{split}
\end{equation}}
$$

$ \Large \displaystyle \int \frac{1}{(1 + x^2)^2} \ dx $

$$
{ \Large \displaystyle
\begin{equation}
\begin{split}
\int \frac{1}{(1 + x^2)^2} \ dx \\
x &= tan \ \theta \\
dx &= \frac{1}{cos^2 \ \theta} d \theta\\
1 + tan^2 \ \theta &= \frac{cos^2 \ \theta}{cos^2 \ \theta} + \frac{sin^2 \ \theta}{cos^2 \ \theta} = \frac{1}{cos^2 \ \theta} \\
\int \frac{1}{(1 + x^2)^2} \ dx &= \int cos^4 \ \theta \frac{1}{cos^2 \ \theta} d \theta\\
&= \int cos^2 \ \theta \ d \theta\\
\end{split}
\end{equation}}
$$

$$
{ \Large \displaystyle
\left( \begin{aligned}
cos^2 \ \theta \ d \theta &= Re \left[ \left( \frac{e^{i \theta} + e^{-i \theta}}{2}\right)^2 \right]\\
&= Re \left[ \frac{e^{2i \theta} +2 e^{i \theta} \cdot e^{-i \theta} + e^{-2i \theta}}{4} \right]\\
&= Re \left[ \frac{e^{2i \theta} + e^{-2i \theta} + 2 }{4} \right]\\
&= \frac{1 + cos (2 \theta) }{2}
\end{aligned} \right) }
$$

$$
{ \Large \displaystyle
\begin{equation}
\begin{split}
\int \frac{1}{(1 + x^2)^2} \ dx &= \int \frac{1 + cos (2 \theta) }{2} \ d \theta\\
&= \frac{1}{2} \theta + \frac{1}{4} sin (2 \theta) + C\\
\end{split}
\end{equation}}
$$

$$
{ \Large \displaystyle
\left( \begin{aligned}
sin (2 \theta) &= \frac{e^{2i \theta}- e^{-2i \theta}}{2i}\\
&= \frac{(e^{i \theta})^2- (e^{-i \theta})^2}{2i} \\
&= \frac{(e^{i \theta} + e^{-i \theta})(e^{i \theta}- e^{-i \theta})}{2i} \\
&= 2\frac{e^{i \theta}- e^{-i \theta}}{2i} \frac{e^{i \theta}+ e^{-i \theta}}{2}\\
&= 2 sin \ \theta \ cos \ \theta \\
\end{aligned} \right) }
$$

$$
{ \Large \displaystyle
\begin{equation}
\begin{split}
\int \frac{1}{(1 + x^2)^2} \ dx &= \frac{1}{2} \theta + \frac{1}{4} sin (2 \theta) + C\\
&= \frac{1}{2} \theta + \frac{1}{2} sin \ \theta \ cos \ \theta + C\\
sin \ \theta \ cos \ \theta &= \frac{sin \ \theta / cos \ \theta}{1/cos^2 \ \theta}
= \frac{sin \ \theta / cos \ \theta}{(sin^2 \ \theta + cos^2 \ \theta)/cos^2 \ \theta} \\
&= \frac{sin \ \theta / cos \ \theta}{(sin \ \theta /cos \ \theta)^2 + (cos \ \theta /cos \ \theta)^2} \\
&= \frac{x}{1 + x^2} \\
\int \frac{1}{(1 + x^2)^2} \ dx &= \frac{1}{2} tan^{-1} x + \frac{x}{2(1 + x^2)} + C\\
\end{split}
\end{equation}}
$$