技巧性的记忆终究归于遗忘，或曰：唯手熟尔。

1. $$ \begin{align} \int\limits_{ -\infty }^{\infty }{\dfrac{1}{{x}^{2}+6x+10}}{\mathrm{d}x} \end{align} $$

The integral: $\int{\dfrac{1}{{x}^{2}+6x+10}}{\mathrm{d}x} \\\\ = \int{\dfrac{1}{\left({x+3}\right)^{2}+1}}{\mathrm{d}x} \\\\ = \int{\dfrac{1}{{u}^{2}+1}}{\mathrm{d}u}; (let: u = x+3) \\\\ =\arctan\left(u\right)+C \\\\ =\arctan\left(x+3\right)+C; (recall: u = x+3) \\\\$

Let: $F\left(x\right)=\arctan\left(x+3\right) \\\\ \lim_{b\to\infty}F\left(b\right)=\dfrac{\pi}{2} \\\\ \lim_{a\to-\infty}F\left(a\right)=-\dfrac{\pi}{2}\\\\$

Thus:
$\int\limits_{ -\infty }^{\infty }{\dfrac{1}{{x}^{2}+6x+10}}{\mathrm{d}x} = \pi$

2. $$ \begin{align} \int\limits_{0}^{\infty }{\dfrac{3}{7\left({x}^{2}+1\right)}}{\mathrm{d}x} \end{align} $$

The integral:

$\int{\dfrac{3}{7\left({x}^{2}+1\right)}}{\mathrm{d}x} \\\\ = \dfrac{3}{7}\int{\dfrac{1}{{x}^{2}+1}}{\mathrm{d}x} \\\\ =\dfrac{3}{7}\cdot\arctan\left(x\right)$

Let: $F\left(x\right)=\dfrac{3\arctan\left(x\right)}{7} \\\\ \lim_{a\to\infty}F\left(a\right)=\dfrac{3\pi}{14} \\\\ F\left(0\right)=0 $

Thus:
$\int\limits_{0}^{\infty }{\dfrac{3}{7\left({x}^{2}+1\right)}}{\mathrm{d}x} = \dfrac{3\pi}{14}$

3. $$ \begin{align} \int\limits_{0}^{5}{\dfrac{1}{\sqrt{25-{x}^{2}}}}{\mathrm{d}x} \end{align} $$

The integral:

$\int{\dfrac{1}{\sqrt{25-{x}^{2}}}}{\mathrm{d}x} \\\\\ = \int{\dfrac{5}{\sqrt{25-25{u}^{2}}}}{\mathrm{d}u ; (let: x = 5u)} \\\\ = \int{\dfrac{1}{\sqrt{1-{u}^{2}}}}{\mathrm{d}u} \\\\ = \arcsin\left(u\right) + C \\\\ = \arcsin\left(\frac{x}{5}\right) + C; (recall: x = 5u)$

Let: $F\left(x\right)=\arcsin\left(\frac{x}{5}\right) \\\\ F\left(0\right)=0 \\\\ \lim_{\delta\to 0}F\left(5-\delta\right)=\dfrac{\pi}{2} \\\\ $

Thus:
$\int\limits_{0}^{5}{\dfrac{1}{\sqrt{25-{x}^{2}}}}{\mathrm{d}x} = \dfrac{\pi}{2} $

4. $$ \begin{align} \int\limits_{-3}^{0}{\dfrac{3}{x+3}}{\mathrm{d}x} \end{align} $$

The integral:

$\int{\dfrac{3}{x+3}}{\mathrm{d}x} \\\\ = 3\int{\dfrac{1}{x+3}}{\mathrm{d}x} \\\\ = 3\int{\dfrac{1}{u}}{\mathrm{d}x ; (let: u = x+3)}\\\\ = 3\ln(\left|u\right|) \\\\ = 3\ln(\left|x+3\right|) $

The integral diverges.

5. $$ \begin{align} \int\limits_{0}^{\infty }{\dfrac{\arctan\left(x\right)}{6\left({x}^{2}+1\right)}}{\mathrm{d}x} \end{align} $$

The integral:

$\int{\dfrac{\arctan\left(x\right)}{6\left({x}^{2}+1\right)}}{\mathrm{d}x} \\\\ = \dfrac{1}{6}\int{\dfrac{\arctan\left(x\right)}{{x}^{2}+1}}{\mathrm{d}x} \\\\ = \dfrac{1}{6}\int{u}{\mathrm{d}u; (let: \begin{gathered}u=\arctan\left(x\right)&\\\mathrm{d}u=\dfrac{1}{{x}^{2}+1}\mathrm{d}x&\end{gathered})} \\\\ =\dfrac{1}{6}\cdot\dfrac{{u}^{2}}{2}\\\\ =\dfrac{\arctan^{2}\left(x\right)}{12}$

Let: $F\left(x\right)=\dfrac{\arctan^{2}\left(x\right)}{12} \\\\ \lim_{a\to\infty}F\left(a\right)=\dfrac{{\pi}^{2}}{48} \\\\ F\left(0\right)=0 \\\\$

Thus:
$\int\limits_{0}^{\infty }{\dfrac{\arctan\left(x\right)}{6\left({x}^{2}+1\right)}}{\mathrm{d}x} = \dfrac{{\pi}^{2}}{48} $

6. $$ \begin{align} \int\limits_{ -\infty }^{0}{{e}^{x}}{\mathrm{d}x} \end{align} $$

The integral:

$\int{{e}^{x}}{\mathrm{d}x} \\\\ = {e}^{x}+C \\\\$

Let: $F\left(x\right)={e}^{x} \\\\ F\left(0\right)=1 \\\\ \lim_{a\to-\infty}F\left(a\right)=0 \\\\$

Thus:
$\int\limits_{ -\infty }^{0}{{e}^{x}}{\mathrm{d}x} = 1 $

7. $$ \begin{align} \int\limits_{0}^{\infty }{\dfrac{12{x}^{2}}{{x}^{3}+1}}{\mathrm{d}x} \end{align} $$

The integral:

$\int{\dfrac{12{x}^{2}}{{x}^{3}+1}}{\mathrm{d}x} \\\\ = 12\int{\dfrac{1}{3u}}{\mathrm{d}u; (let: \begin{gathered}u={x}^{3}+1&\\\dfrac{1}{3}\mathrm{d}u={x}^{2}\mathrm{d}x&\end{gathered})} \\\\ = 4\ln\left(\left|u\right|\right) \\\\ = 4\ln\left(\left|x^3+1\right|\right) $

The integral diverges.

8. $$ \begin{align} \int\limits_{0}^{3}{\dfrac{1}{6-2x}}{\mathrm{d}x} \end{align} $$

The integral:

$\int{\dfrac{1}{6-2x}}{\mathrm{d}x} \\\\ = \int{-\dfrac{1}{2\left(x-3\right)}}{\mathrm{d}x} \\\\ = -\dfrac{1}{2}\int{\dfrac{1}{x-3}}{\mathrm{d}x} \\\\ = -\dfrac{\ln\left(\left|u\right|\right)}{2} \\\\ = -\dfrac{\ln\left(\left|x-3\right|\right)}{2}$

9. $$ \begin{align} \int{\dfrac{\sqrt[{x}]{e}}{{x}^{e}}}{\mathrm{d}x} \end{align} $$

The integral: $\int\frac{\sqrt[x]{e}}{x^e} d x \\\\ =-\left(-\frac{1}{x}\right)^{-e} x^{-e} \Gamma\left(-1+e,-\frac{1}{x}\right) + C$

10. $$ \begin{align} \int\limits_{0}^{1}{\ln\left(4x\right)}{;\mathrm{d}x} \end{align} $$

The integral:

$\int{\ln\left(4x\right)}{\;\mathrm{d}x} \\\\ = \int{\dfrac{\ln\left(u\right)}{4}}{\;\mathrm{d}u ; (let: \def\arraystretch{2}\begin{array}{c|c}u=4x&x=\dfrac{u}{4}\\&\mathrm{d}x=\dfrac{1}{4}\mathrm{d}u\end{array})} \\\\ = \dfrac{1}{4}\left(u\ln\left(u\right)-\int{1}{\;\mathrm{d}u}\right)\\\\ = \dfrac{u\ln\left(u\right)}{4}-\dfrac{u}{4} \\\\ = x\ln\left(4x\right)-x$

Thus:
$\int\limits_{0}^{1}{\ln\left(4x\right)}{\;\mathrm{d}x} = \ln\left(4\right)-1 $