与えられた不定積分を計算する問題です。 2) $\int \frac{3x+2}{3x^2+6x+4} dx$ 3) $\int \frac{x \sin^{-1}x}{(1-x^2)^{3/2}} dx$ 4) $\int 2e^x \cos^2x dx$ 5) $\int \tan^{-1}(3x) dx$

1. 問題の内容

与えられた不定積分を計算する問題です。

2) $\int \frac{3x+2}{3x^2+6x+4} dx$

3) $\int \frac{x \sin^{-1}x}{(1-x^2)^{3/2}} dx$

4) $\int 2e^x \cos^2x dx$

5) $\int \tan^{-1}(3x) dx$

2. 解き方の手順

2) $\int \frac{3x+2}{3x^2+6x+4} dx$

分母を $u = 3x^2+6x+4$ とおくと、$du = (6x+6)dx = 2(3x+3)dx$ となります。

よって、与式は

$\int \frac{3x+2}{3x^2+6x+4} dx = \int \frac{3x+3 - 1}{3x^2+6x+4} dx = \int \frac{3x+3}{3x^2+6x+4} dx - \int \frac{1}{3x^2+6x+4} dx$

$= \frac{1}{2} \int \frac{2(3x+3)}{3x^2+6x+4} dx - \int \frac{1}{3(x^2+2x)+4} dx$

$= \frac{1}{2} \int \frac{du}{u} - \int \frac{1}{3(x^2+2x+1) + 4 - 3} dx$

$= \frac{1}{2} \ln|u| - \int \frac{1}{3(x+1)^2+1} dx$

$= \frac{1}{2} \ln|3x^2+6x+4| - \frac{1}{3} \int \frac{1}{(x+1)^2+\frac{1}{3}} dx$

$v = x+1$ とすると、$dv = dx$ なので、

$= \frac{1}{2} \ln|3x^2+6x+4| - \frac{1}{3} \int \frac{1}{v^2+\frac{1}{3}} dv$

$= \frac{1}{2} \ln|3x^2+6x+4| - \frac{1}{3} \cdot \frac{1}{\sqrt{\frac{1}{3}}} \tan^{-1}\left(\frac{v}{\sqrt{\frac{1}{3}}}\right) + C$

$= \frac{1}{2} \ln|3x^2+6x+4| - \frac{1}{\sqrt{3}} \tan^{-1}(\sqrt{3}(x+1)) + C$

3) $\int \frac{x \sin^{-1}x}{(1-x^2)^{3/2}} dx$

$\sin^{-1}x = t$ とおくと、$x = \sin t$、$dx = \cos t dt$

$\int \frac{x \sin^{-1}x}{(1-x^2)^{3/2}} dx = \int \frac{\sin t \cdot t}{(1-\sin^2t)^{3/2}} \cos t dt = \int \frac{\sin t \cdot t}{(\cos^2t)^{3/2}} \cos t dt = \int \frac{\sin t \cdot t}{\cos^3t} \cos t dt = \int \frac{\sin t}{\cos^2t} t dt$

$u = \frac{1}{\cos t}$, $dv = t \sin t dt$, $du = \frac{\sin t}{\cos^2 t}dt$, $v = t(-\cos t) - \int (-\cos t) dt = -t\cos t + \sin t$

$\int \frac{\sin t}{\cos^2t} t dt = \int t (\cos t)^{-2} \sin t dt = ut - \int u' t dx$

Let $u = \sin^{-1} x$. Then $x = \sin u$, $dx = \cos u du$.

$\int \frac{x \sin^{-1}x}{(1-x^2)^{3/2}} dx = \int \frac{\sin u \cdot u}{(1-\sin^2 u)^{3/2}} \cos u du = \int \frac{\sin u \cdot u}{\cos^3 u} \cos u du = \int \frac{u \sin u}{\cos^2 u} du$.

Let $v = u$, $dw = \frac{\sin u}{\cos^2 u} du$. Then $dv = du$, $w = \frac{1}{\cos u}$.

$\int \frac{u \sin u}{\cos^2 u} du = \frac{u}{\cos u} - \int \frac{1}{\cos u} du = \frac{\sin^{-1} x}{\sqrt{1-x^2}} - \int \frac{\cos u}{\cos^2 u} du = \frac{\sin^{-1} x}{\sqrt{1-x^2}} - \int \frac{\cos u}{1-\sin^2 u} du = \frac{\sin^{-1} x}{\sqrt{1-x^2}} - \int \frac{\cos u}{1-x^2} du$

$= \frac{\sin^{-1} x}{\sqrt{1-x^2}} - \int \frac{1}{1-x^2} dx = \frac{\sin^{-1} x}{\sqrt{1-x^2}} - \frac{1}{2} \ln|\frac{1+x}{1-x}| + C$

4) $\int 2e^x \cos^2x dx$

$\cos^2 x = \frac{1 + \cos 2x}{2}$ より、

$\int 2e^x \cos^2x dx = \int 2e^x \frac{1 + \cos 2x}{2} dx = \int e^x (1+\cos 2x) dx = \int e^x dx + \int e^x \cos 2x dx$

$\int e^x \cos 2x dx = e^x \cos 2x - \int e^x (-2\sin 2x) dx = e^x \cos 2x + 2\int e^x \sin 2x dx$

$= e^x \cos 2x + 2(e^x \sin 2x - \int e^x (2\cos 2x) dx) = e^x \cos 2x + 2e^x \sin 2x - 4 \int e^x \cos 2x dx$

$5 \int e^x \cos 2x dx = e^x \cos 2x + 2e^x \sin 2x$

$\int e^x \cos 2x dx = \frac{1}{5} e^x (\cos 2x + 2\sin 2x)$

$\int e^x (1+\cos 2x) dx = e^x + \frac{1}{5} e^x (\cos 2x + 2\sin 2x) + C = e^x (1 + \frac{1}{5} \cos 2x + \frac{2}{5} \sin 2x) + C$

5) $\int \tan^{-1}(3x) dx$

$u = \tan^{-1}(3x)$, $dv = dx$. Then $du = \frac{3}{1+(3x)^2} dx = \frac{3}{1+9x^2} dx$, $v = x$.

$\int \tan^{-1}(3x) dx = x\tan^{-1}(3x) - \int \frac{3x}{1+9x^2} dx = x\tan^{-1}(3x) - \frac{1}{6} \int \frac{18x}{1+9x^2} dx$

$w = 1+9x^2$, $dw = 18x dx$

$= x\tan^{-1}(3x) - \frac{1}{6} \int \frac{1}{w} dw = x\tan^{-1}(3x) - \frac{1}{6} \ln|w| + C = x\tan^{-1}(3x) - \frac{1}{6} \ln|1+9x^2| + C$

3. 最終的な答え

2) $\frac{1}{2} \ln|3x^2+6x+4| - \frac{1}{\sqrt{3}} \tan^{-1}(\sqrt{3}(x+1)) + C$

3) $\frac{\sin^{-1} x}{\sqrt{1-x^2}} - \frac{1}{2} \ln|\frac{1+x}{1-x}| + C$

4) $e^x (1 + \frac{1}{5} \cos 2x + \frac{2}{5} \sin 2x) + C$

5) $x\tan^{-1}(3x) - \frac{1}{6} \ln(1+9x^2) + C$