问题及解答

求不定积分

\[\int\ln(\sqrt{1+x}+\sqrt{1-x})dx.\]

\[
\begin{split}
\int\ln(\sqrt{1+x}+\sqrt{1-x})dx&=\int\frac{1}{2}\ln(\sqrt{1+x}+\sqrt{1-x})^2dx\\
&=\frac{1}{2}\int\ln(2+2\sqrt{1-x^2})dx\\
&=\frac{1}{2}\int\ln 2dx+\frac{1}{2}\int\ln(1+\sqrt{1-x^2})dx\\
&=\frac{x}{2}\ln 2+\frac{1}{2}\int\ln(1+\sqrt{1-x^2})dx.
\end{split}
\]

对于不定积分, 我们令 $x=\sin\theta$, 其中 $\theta\in[0,\frac{\pi}{2})$, 于是

\[
\begin{split}
\int\ln(1+\sqrt{1-x^2})dx&=\int\ln(1+\cos\theta)d\sin\theta\\
&=\sin\theta\ln(1+\cos\theta)-\int\sin\theta d\ln(1+\cos\theta)\\
&=\sin\theta\ln(1+\cos\theta)-\int\sin\theta\cdot\frac{-\sin\theta}{1+\cos\theta}d\theta\\
&=x\ln(1+\sqrt{1-x^2})+\int\frac{\sin^2 \theta}{1+\cos\theta}d\theta,\\
\end{split}
\]

而

\[
\int\frac{\sin^2 \theta}{1+\cos\theta}d\theta=\int\frac{1-\cos^2\theta}{1+\cos\theta}d\theta=\int(1-\cos\theta)d\theta=\theta-\sin\theta+C=\arcsin x-x+C.
\]

因此原不定积分等于

\[
\begin{split}
\int\ln(\sqrt{1+x}+\sqrt{1-x})dx&=\frac{x}{2}\ln 2+\frac{1}{2}\cdot x\ln(1+\sqrt{1-x^2})+\frac{1}{2}(\arcsin x-x)+C\\
&=\frac{x}{2}(\ln 2+\ln(1+\sqrt{1-x^2}))+\frac{1}{2}(\arcsin x-x)+C\\
&=x\ln(\sqrt{1+x}+\sqrt{1-x})+\frac{1}{2}(\arcsin x-x)+C.
\end{split}
\]