U-substitutions

If $u = x + 3$, then $$du = dx.$$ And so, $$ \int (x + 3)^2 \, dx = \int u^2 \, du,$$ which is readily solved:
$$\int u^2 \, du = \dfrac{1}{3}u^3 + C,$$ where $C$ is an arbitrary constant.

However, we have to write the answer in terms of $x$ and not $u$, so we undo the substitution we did at the beginning of the problem to get \begin{eqnarray} \int (x+3)^2\, dx &=& \int u^2 \, du\\ \ &=& \dfrac{1}{3}u^3 + C \\ \ &=& \dfrac{1}{3}(x + 3)^3 + C, \end{eqnarray} for some arbitrary constant $C$.

Thinking: The power rule for integration is $$\int x^n\, dx = \dfrac{x^{n+1}}{n+1} + C, \text{ for } n \neq -1$$ and so is not applicable here. So, we need another method.
For u-substitution, we can't really set $u = x$, since that won't change anything. Instead, we notice that if we set $x = e^u$, then its differential, $dx = e^u\, du$, is proportional to $x$ and would likely simplify the problem.

Solution: Let $x = e^u$, then $$dx = e^u\, du$$ and so \begin{eqnarray} \int\dfrac{dx}{x} &=& \int \dfrac{e^u\, du}{e^u} \\ \ &=& \int du \\ \ &=& u + C \\ \ &=& \ln (x) + C, \end{eqnarray} for an arbitrary constant $C$.

In the last step, we made use of the fact that if $x = e^u$, then $u = \ln (x)$.

Thinking: Since the numerator ($\cos x\, dx$) is the differential of the denominator ($\sin x$), we set the denominator to be $u$. This should simplify the problem.

Solution: If $u = \sin x$, then $$du = \cos x \, dx.$$ And so, $$\int \cot x\, dx = \int \dfrac{\cos x }{\sin x} \, dx = \int \dfrac{du}{u}.$$

Then, recall that the derivative of $\ln (x)$ with respect to $x$ is $1/x$, that is, $$\dfrac{d}{dx}\ln(x) = \dfrac{1}{x}.$$ (If you don't remember this, check out the previous question to see why this is true.)
And so, $$\int \dfrac{dx}{x} = \ln(x) + C,$$ for some arbitrary constant $C$.

We then apply this result back to our original problem: \begin{eqnarray} \int \dfrac{\cos x }{\sin x} \, dx &=& \int \dfrac{du}{u} \\ \ &=& \ln(u) + C \\ \ &=& \ln (\sin x) + C, \end{eqnarray} for some arbitrary constant $C$.

Thinking: As with the previous problem, notice that the numerator is proportional to the denominator. That is, $$d (\cos x) = -\sin x\, dx.$$ So, setting $u$ to the denominator should simplify the problem.

Solution: Let $u = \cos x$, then $du = -\sin x\, dx$. And so, \begin{eqnarray} \int \tan x\, dx &=& \int \dfrac{\sin x\, dx}{\cos x} \\ \ &=& -\int \dfrac{du}{u} \\ \ &=& -\ln (u) + C \\ \ &=& -\ln (\cos x) + C, \end{eqnarray} for some arbitrary constant $C$.

Thinking: The numerator is proportional to the differential of the denominator, so setting $u$ to equal the denominator should simplify the problem.

Solution: Let $u = 1 + 3x^2$, then $$du = 6x\, dx,$$ or $$ x\, dx = \dfrac{du}{6}.$$ And so, \begin{eqnarray} \int \dfrac{x}{1 + 3x^2}\, dx &=& \int \dfrac{(x\, dx)}{(1 + 3x^2)} \\ \ &=& \dfrac{1}{6} \int \dfrac{du}{u} \\ \ &=& \dfrac{1}{6} \ln (u) + C \\ \ &=& \dfrac{1}{6} \ln (1 + 3x^2) + C, \end{eqnarray} for some arbitrary constant $C$.

If $u = 1 + 3x^2$, then $$x^2 = \dfrac{1}{3}(u - 1)$$ and $$du = 6x\, dx.$$ We may also solve for $dx$ to get \begin{eqnarray} dx &=& \dfrac{du}{6x} \\ \ &=& \dfrac{du}{6 \sqrt{\dfrac{1}{3}(u - 1)}} \end{eqnarray} $$ $$ And so, \begin{eqnarray} \int x^2 (1 + 3 x^2) \, dx &=& \int \dfrac{u - 1}{3} \cdot u \cdot \dfrac{du}{6 \sqrt{\dfrac{1}{3}(u - 1)}} \\ \ &=& \dfrac{\sqrt{3}}{18} \int \dfrac{u \cdot (u - 1) \, du}{\sqrt{u-1}} \\ \ &=& \dfrac{\sqrt{3}}{18} \int u \cdot \sqrt{u - 1} \, du. \end{eqnarray}

Let $$u = x^2 + 4,$$ then $$du = 2x\, dx.$$ And so, \begin{eqnarray} \int 3x \cdot (x^2 + 4)^{13}\, dx &=& \dfrac{3}{2} \int u^{13}\, du \\ \ &=& \dfrac{3}{8}u^{14} + C, \end{eqnarray} where $C$ is an arbitrary constant. $$ $$ Next, we simply need to rewrite the answer in terms of $x$ (we do this by plugging back in the $u = x^2 + 4$). The answer is therefore $$\int 3x \cdot (x^2 + 4)^{13}\, dx = \dfrac{3}{8}(x^2 + 4)^{14} + C,$$ where $C$ is an arbitrary constant.

Let $$u = x^2 + 3x,$$ then $$du = (2x + 3)\, dx.$$ And so, \begin{eqnarray} \int \dfrac{2x + 3}{(x^2 + 3x)^2}\, dx &=& \int \dfrac{du}{u^2} \\ \ &=& \int u^{-2}\, du \\ \ &=& \dfrac{u^{-1}}{-1} + C \\ \ &=& \dfrac{-1}{x^2 + 3x} + C, \end{eqnarray} where $C$ is an arbitrary constant.