\(\displaystyle \int e^{x + e^x}\,dx\)
\(\begin{align}
& \int e^{x + e^x}\,dx \\[0.4em]
=\ & \int e^x \cdot e^{e^x}\,dx \\[0.4em]
=\ & \int e^u\,du \\[0.4em]
=\ & e^u + C \\[0.4em]
=\ & e^{e^x} + C
\end{align}\)

\(\begin{align}
& u = e^x \\
& du = e^x\,dx
\end{align}\)

\(\displaystyle \int \cot x[\ln(\sin x)]\,dx\)
\(\begin{align}
& \int \cot x[\ln(\sin x)]\,dx \\[0.4em]
=\ & \int u\,du \\[0.4em]
=\ & \dfrac{u^2}{2} + C \\[0.4em]
=\ & \dfrac{[\ln(\sin x)]^2}{2} + C
\end{align}\)

\(\begin{align}
& u = \ln(\sin x) \\[0.4em]
& du = \dfrac{1}{\sin x}\cos x\,dx \\[0.4em]
& du = \cot x\,dx
\end{align}\)

\(\displaystyle \int \dfrac{x}{1 + x^4}\,dx\)
\(\begin{align}
& \int \dfrac{x}{1 + x^4}\,dx \\[0.75em]
=\ & \int \dfrac{x}{1 + \left(x^2\right)^2}\,dx \\[0.75em]
=\ & \int \dfrac{x}{1 + u^2} \cdot \dfrac{du}{2x} \\[0.75em]
=\ & \int \dfrac{1}{1 + u^2} \cdot \dfrac{du}{2} \\[0.75em]
=\ & \dfrac{1}{2} \int \dfrac{1}{1 + u^2}\,du \\[0.75em]
=\ & \dfrac{1}{2}\tan^{1} u + C \\[0.75em]
=\ & \dfrac{1}{2}\tan^{1}\left(x^2\right) + C
\end{align}\)

\(\begin{align}
& u = x^2 \\
& du = 2x\,dx
\end{align}\)

\(\displaystyle \int \tan x\ln(\cos x)\,dx\)
\(\begin{align}
& \int \tan x\ln(\cos x)\,dx \\[0.75em]
=\ & \int (\tan x) u\dfrac{du}{\tan x} \\[0.75em]
=\ & \int u\,du \\[0.75em]
=\ & \dfrac{1}{2}u^2 + C \\[0.75em]
=\ & \dfrac{1}{2}[\ln(\cos x)]^2 + C
\end{align}\)

\(\begin{align}
& u = \ln(\cos x) \\[0.4em]
& du = \dfrac{1}{\cos x} \cdot (\sin x)\,dx \\[0.4em]
& du = \tan x\,dx \\[0.4em]
& \dfrac{du}{\tan x} = dx
\end{align}\)

\(\displaystyle \int \dfrac{1}{1 + \sqrt{x}}\,dx\)
\(\begin{align}
& \int \dfrac{1}{1 + \sqrt{x}}\,dx \\[0.75em]
=\ & \int \dfrac{1}{u}2(u  1)\,du \\[0.75em]
=\ & 2\int \dfrac{u  1}{u}\,du \\[0.75em]
=\ & 2\int \dfrac{u}{u}  \dfrac{1}{u}\,du \\[0.75em]
=\ & 2\int 1  \dfrac{1}{u}\,du \\[0.75em]
=\ & 2(u  \ln \lvert u \rvert + C) \\[0.75em]
=\ & 2u  2\ln \lvert u \rvert + C \\[0.75em]
=\ & 2(1 + \sqrt{x})  2\ln \lvert 1 + \sqrt{x} \rvert + C \\[0.75em]
=\ & 2\sqrt{x}  2\ln(1 + \sqrt{x}) + C
\end{align}\)

\(\begin{align}
u = 1 + \sqrt{x} \\[0.4em]
u  1 = \sqrt{x} \\[0.4em]
du = \dfrac{1}{2\sqrt{x}}\,dx \\[0.4em]
2(u  1)\,du = dx
\end{align}\)

\(\displaystyle \int \dfrac{e^x + 1}{e^x}\,dx\)
\(x  e^{x} + C\)
\(\displaystyle \int 3^{\sin \theta}\cos \theta\,d\theta\)
\(\dfrac{3^{\sin \theta}}{\ln 3} + C\)