[Ne[S]zA]

$$\int \dfrac{1}{2^x+2^{-x}+\sqrt{2}}dx$$
ให้ $u=2^x$ ได้ว่า $du = 2^x \ln 2 dx$
ได้ว่า
$$\int \dfrac{1}{2^x+2^{-x}+\sqrt{2}} \times \dfrac{du}{2^x \ln 2}=\dfrac{1}{\ln 2}\int \dfrac{1}{2^{2x}+\sqrt{2}(2^x)+1}du=\dfrac{1}{\ln 2}\int \dfrac{1}{u^2+\sqrt{2}u+1}du=\dfrac{1}{\ln 2}\int \dfrac{1}{(u+\frac{1}{\sqrt{2}})^2+(\frac{1}{\sqrt{2}})^2}du$$
ให้ $u+\frac{1}{\sqrt{2}}=\frac{1}{\sqrt{2}}\tan \theta$ ได้ว่า $\sqrt{2}u+1=\tan \theta$ เพราะฉะนั้น $\theta = \tan^{-1}(\sqrt{2}u+1)$ และ $u=\frac{1}{\sqrt{2}}(\tan \theta -1)$ ได้ว่า $du=\frac{1}{\sqrt{2}}\sec^2\theta d\theta$
จึงได้ว่า
$$\dfrac{1}{\ln 2}\int \dfrac{2}{\tan^2\theta+1} \times \dfrac{\sec^2\theta d\theta}{\sqrt{2}}=\dfrac{\sqrt{2}}{\ln 2}\int d\theta=\dfrac{\sqrt{2}}{\ln 2} \times \theta+C=\dfrac{\sqrt{2}}{\ln 2} [\tan^{-1}(\sqrt{2}u+1)]+C=\dfrac{\sqrt{2}}{\ln 2}[\tan^{-1}(\sqrt{2}(2^x)+1)]+C$$
$$\therefore \int \dfrac{1}{2^x+2^{-x}+\sqrt{2}}dx=\dfrac{\sqrt{2}}{\ln 2}[\tan^{-1}(2^{x+\frac{1}{2}}+1)]+C$$