Let $$u =\answer [given]{\ln (x)},$$ computing $du$, we find $$du =\answer [given]{\frac {1}{x}} dx$$ and solving for $dx$ we find $$dx = \answer [given]{x} du.$$ Now $$\begin{align*} \int_{2}^{3} \frac{1}{x\ln(x)} dx &= \int_{g(2)}^{g(3)} \frac{1}{x \cdot u} x du \\ &= \int_{\answer[given]{\ln(2)}}^{\answer[given]{\ln(3)}} \answer[given]{\frac{1}{u}} du\\ &= \bigg[ \answer[given]{\ln(u)} \bigg]_{\answer[given]{\ln(2)}}^{\answer[given]{\ln(3)}}\\ & = \ln(\ln(3)) - \ln(\ln(2)). \end{align*}$$

Here it is not apparent that the chain rule is involved. However, if it was involved, perhaps a good guess for $g$ would be $$u = \answer [given]{1-x^2}$$ and then $$\begin{align*} du &= \answer[given]{-2x} dx, \\ dx &= \answer[given]{\frac{-1}{2x}} du. \end{align*}$$ Now we consider the integral we are trying to compute $$\int x^3\sqrt {1-x^2} dx$$ and we substitute using our work above. Write with me $$\begin{align*} \int x^3\sqrt{1-x^2} dx &= \int x^3 \answer[given]{\sqrt{u}} \left( \answer[given]{\frac{-1}{2x}} \right) du \\ &= \int \frac{-x^2 \answer[given]{\sqrt{u}}}{2} du. \end{align*}$$ However, we cannot continue until each $x$ is replaced. We know that $$\begin{align*} u &= 1-x^2 \\ \Rightarrow \qquad u -1 &= -x^2\\ \Rightarrow \qquad \answer[given]{1- u} &= x^2 \end{align*}$$ so now we may replace $x^2$ $$\int x^3\sqrt {1-x^2} dx = \int -\frac {\answer [given]{(1-u)} \sqrt {u}}{2} du.$$ At this point, we are close to being done. Write $$\begin{align*} \int -\frac{(1-u)\sqrt{u}}{2} du &= \int \left(\frac{u\sqrt{u}}{2} - \frac{\sqrt{u}}{2}\right) du \\ &= \int \frac{\answer[given]{u^{3/2}}}{2} du - \int \frac{\sqrt{u}}{2} du \\ &= \answer[given]{\frac{u^{5/2}}{5}} - \answer[given]{\frac{u^{3/2}}{3}}. \end{align*}$$ Now recall that $u = 1-x^2$. Hence our final answer is $$\int x^3\sqrt {1-x^2} dx = \answer [given]{\frac {(1-x^2)^{5/2}}{5}} - \frac {(1-x^2)^{3/2}}{3}+C.$$

We substitute $$u = \sec (y) + \tan (y)$$ and we immediately see that $$\begin{align*} du &= \answer[given]{(\sec(y) \tan(y) + \sec^2(y))} dy,\\ dy &= \frac{1}{\answer[given]{\sec(y) \tan(y) + \sec^2(y)}} du. \end{align*}$$ But this cancels perfectly with the numerator! So we have that $$\begin{align*} \int \frac{\sec(y) \tan(y) + \sec^2(y)}{\sec(y) + \tan(y)} dy &= \int \frac{1}{u} du \\ &= \ln|u| + C \\ &= \ln| \sec(y) + \tan(y) | + C. \end{align*}$$

We want to substitute for $1-u^2$. But the variable “$u$” has already been used…OH NO! Never fear! We can substitute with whatever variable that we want. In particular, let us use “$w$” for this problem. So we let $$w = 1 - u^2$$ and then $$\begin{align*} dw &= \answer[given]{-2u} du,\\ du &= \answer[given]{\frac{-1}{2u}} dw. \end{align*}$$ Thus $$\begin{align*} \int \frac{u}{1-u^2} du &= \int \frac{u}{w} \left( \answer[given]{\frac{-1}{2u}} \right) dw \\ &= - \frac{1}{2} \int \frac{1}{w} dw \\ &= - \frac{1}{2} \ln|w| + C \\ &= - \frac{1}{2} \ln|1-u^2| + C. \end{align*}$$

We begin by writing $$\begin{align*} \tan(x) &= \frac{\sin(x)}{\cos(x)} \\ &= \frac{\sin(x)}{\cos(x)} \cdot \frac{\cos(x)}{\cos(x)} \\ &= \frac{\sin(x) \cos(x)}{1 - \sin^2(x)}. \end{align*}$$ We then make the substitution $$s = \sin (x)$$ and so $$\begin{align*} ds &= \answer[given]{\cos(x)} dx,\\ dx &= \frac{1}{\answer[given]{\cos(x)}} ds. \end{align*}$$ Then $$\begin{align*} \int \tan(x) dx &= \int \frac{\sin(x) \cos(x)}{1 - \sin^2(x)} dx \\ &= \int \frac{s \cos(x)}{1 - s^2} \cdot \frac{1}{\answer[given]{\cos(x)}} ds \\ &= \int \frac{s}{1-s^2} ds. \end{align*}$$ But this is the same problem as Example key example! And so we know that $$\begin{align*} \int \tan(x) dx &= - \frac{1}{2} \ln|1-s^2| + C \\ &= - \frac{1}{2} \ln|1-\sin^2 (x)| + C \\ &= - \frac{1}{2} \ln|\cos^2(x)| + C \\ &= \ln|\cos^2(x)|^{- \frac{1}{2}} + C \\ &= \ln|\sec(x)| + C. \end{align*}$$

Maybe the biggest key to solving this problem is to recall that $$e^{2x} = (\answer [given]{e^x})^2.$$ So we can rewrite the problem $$\int \frac {e^{2x}}{1 - e^{2x}} dx = \int \frac {(\answer [given]{e^x})^2}{1 - (\answer [given]{e^x})^2} dx.$$ Now, if we make the substitution $u = e^x$, we have that $$\begin{align*} du &= \answer[given]{e^x} dx,\\ dx &= \frac{1}{\answer[given]{e^x}} du, \end{align*}$$ and $$\begin{align*} \int \frac{e^{2x}}{1 - e^{2x}} dx &= \int \frac{(e^x)^2}{1 - u^2} \cdot \frac{1}{\answer[given]{e^x}} du \\ &= \int \frac{e^x}{1-u^2} du \\ &= \int \frac{u}{1-u^2} du. \end{align*}$$ But now we are back to Example key example, and so we know that $$\begin{align*} \int \frac{e^{2x}}{1 - e^{2x}} dx &= - \frac{1}{2} \ln|1-u^2| + C \\ &= - \frac{1}{2} \ln|1 - e^{2x}| + C. \end{align*}$$

While it is not obvious at all, let us try the substitution $$u = \sqrt {x}.$$ Then $$\begin{align*} du &= \answer[given]{\frac{1}{2 \sqrt{x}}} dx,\\ dx &= 2 \sqrt{x} du = 2u du, \end{align*}$$ and so $$\begin{align*} \int_0^{16} \sqrt{4 - \sqrt{x}} dx &= \int_{g(0)}^{g(16)} \answer[given]{\sqrt{4-u} \cdot 2u} du \\ &= \int_{\answer[given]{0}}^{\answer[given]{4}} \answer[given]{2u \sqrt{4-u}} du. \end{align*}$$ From here we now make the second (and more obvious) substitution $$w = 4-u.$$ Then $u = 4-w$, and $$\begin{align*} dw &= - du,\\ du &= - dw. \end{align*}$$ So $$\begin{align*} \int_0^{16} \sqrt{4 - \sqrt{x}} dx &= \int_0^4 2u \sqrt{4-u} du \\ &= \int_{w(0)}^{w(4)} 2 (4-w) \sqrt{w} (-1) dw \\ &= - \int_{\answer[given]{4}}^{\answer[given]{0}} ( 8w^{\frac{1}{2}} - 2w^{\frac{3}{2}} ) dw \\ &= \int_{\answer[given]{0}}^{\answer[given]{4}} ( 8w^{\frac{1}{2}} - 2w^{\frac{3}{2}} ) dw \\ &= \bigg[ \answer[given]{8 \left( \frac{2}{3} \right) w^{\frac{3}{2}} - 2 \left( \frac{2}{5} \right) w^{\frac{5}{2}}} \bigg]_{0}^{4} \\ &= \left( \frac{16}{3} (4)^{\frac{3}{2}} - \frac{4}{5} (4)^{\frac{5}{2}} \right) - \left( 0 - 0 \right) \\ &= \frac{128}{3} - \frac{128}{5} \\ &= \frac{256}{15}. \end{align*}$$