$\displaystyle \int \dfrac{x+2}{x^2+4x-13}\,\mathrm{d}x$

$\displaystyle \int \dfrac{x+2}{x^2+4x-13}\,\mathrm{d}x= \dfrac{1}{2}\int \dfrac{2x+4}{x^2+4x-13}\,\mathrm{d}x \,\overset{\substack{t=x^2+4x-13\\ \mathrm{d}t=(2x+4)\,\mathrm{d}x\\ \phantom{\downarrow}}}{=}\, \dfrac{1}{2}\int \dfrac{1}{t}\,\mathrm{d}t= \dfrac{1}{2}\ln|t| + k \,\overset{\substack{t=x^2+4x-13\\ \phantom{\downarrow}}}{=}\, \bbox[#FFECB3,5px]{\dfrac{1}{2}\ln|x^2+4x-13| + k}$

$\displaystyle \int e^x\sin(e^x)\,\mathrm{d}x$

$\displaystyle \int e^x\sin(e^x)\,\mathrm{d}x \,\overset{\substack{t=e^x\\ \mathrm{d}t=e^x\,\mathrm{d}x\\ \phantom{\downarrow}}}{=}\, \int \sin t\,\mathrm{d}t= -\cos t + k \,\overset{\substack{t=e^x\\ \phantom{\downarrow}}}{=}\, \bbox[#FFECB3,5px]{-\cos(e^x) + k}$

$\displaystyle \int \dfrac{\ln(\sqrt{x})}{x}\,\mathrm{d}x$

$\displaystyle \int \dfrac{\ln(\sqrt{x})}{x}\,\mathrm{d}x= \dfrac{1}{2}\int \dfrac{\ln x}{x}\,\mathrm{d}x \,\overset{\substack{t=\ln x\\ \mathrm{d}t=\frac{1}{x}\,\mathrm{d}x\\ \phantom{\downarrow}}}{=}\, \dfrac{1}{2}\int t\,\mathrm{d}t= \dfrac{1}{4}t^2 + k \,\overset{\substack{t=\ln x\\ \phantom{\downarrow}}}{=}\, \bbox[#FFECB3,5px]{\dfrac{1}{4}(\ln x)^2 + k}$

$\displaystyle \int \ln(\cos x)\tan x\,\mathrm{d}x$

$\displaystyle \int \ln(\cos x)\tan x\,\mathrm{d}x= \int \ln(\cos x)\dfrac{\sin x}{\cos x}\,\mathrm{d}x \,\overset{\substack{t=\cos x\\ \mathrm{d}t=-\sin x\,\mathrm{d}x\\ \phantom{\downarrow}}}{=}\, -\int \dfrac{\ln t}{t}\,\mathrm{d}t \,\overset{\substack{w=\ln t\\ \mathrm{d}w=\frac{1}{t}\,\mathrm{d}t\\ \phantom{\downarrow}}}{=}\,\\ \qquad\displaystyle -\int w\,\mathrm{d}w= -\dfrac{1}{2}w^2+k \,\overset{\substack{w=\ln t\\ \phantom{\downarrow}}}{=}\, -\dfrac{1}{2}(\ln t)^2+k \,\overset{\substack{t=\cos x\\ \phantom{\downarrow}}}{=}\, \bbox[#FFECB3,5px]{-\dfrac{1}{2}(\ln(\cos x))^2+k}$

$\displaystyle \int x\sqrt{1-3x^2}\,\mathrm{d}x$

$\displaystyle \int x\sqrt{1-3x^2}\,\mathrm{d}x= -\dfrac{1}{6}\int \sqrt{1-3x^2}(-6x)\,\mathrm{d}x \,\overset{\substack{t=1-3x^2\\ \mathrm{d}t=-6x\,\mathrm{d}x\\ \phantom{\downarrow}}}{=}\, -\dfrac{1}{6}\int t^{\frac{1}{2}}\,\mathrm{d}t= -\dfrac{1}{6}\dfrac{2}{3}t^{\frac{3}{2}} + k \,\overset{\substack{t=1-3x^2\\ \phantom{\downarrow}}}{=}\, \bbox[#FFECB3,5px]{-\dfrac{1}{9}(1-3x^2)^{\frac{3}{2}} + k}$