I know that it would be easier for you if the videos had explanations but I prefer to let you think why I am doing each step and if you don't understand it you can ask in a comment. In my opinion, it is the best way to practice integration.

When we have a polynomial division where the degree of the polynomial on the numerator and denominator is equal, sometimes we can add and subtract a constant in order to avoid performing the long polynomial division. It is a little trick that makes the integral easier and faster to solve. Here you have some examples to start practicing it:

$$\int \frac{x}{x+1} \hspace{0.1cm} dx = $$

$$= \int \frac{x+1-1}{x+1} \hspace{0.1cm} dx = $$

$$= \int \left( \frac{x+1}{x+1} - \frac{1}{x+1} \right) \hspace{0.1cm} dx = $$

$$= \int \left( 1 - \frac{1}{x+1} \right) \hspace{0.1cm} dx = $$

$$= \int 1 \hspace{0.1cm} dx - \int \frac{1}{x+1} \hspace{0.1cm} dx = $$

$$= x - \ln|x+1| + C $$

$$\int \frac{x^2}{x^2+1} \hspace{0.1cm} dx = $$

$$= \int \frac{x^2+1-1}{x^2+1} \hspace{0.1cm} dx = $$

$$= \int \left( \frac{x^2+1}{x^2+1} - \frac{1}{x^2+1} \right) \hspace{0.1cm} dx = $$

$$= \int \left( 1 - \frac{1}{x^2+1} \right) \hspace{0.1cm} dx = $$

$$= \int 1 \hspace{0.1cm} dx - \int \frac{1}{x^2+1} \hspace{0.1cm} dx = $$

$$= x - \arctan(x) + C $$

$$\int \frac{2x}{3x+7} \hspace{0.1cm} dx = $$

$$= 2 \cdot \int \frac{x}{3x+7} \hspace{0.1cm} dx = $$

$$= \frac{2}{3} \cdot \int \frac{3x}{3x+7} \hspace{0.1cm} dx = $$

$$= \frac{2}{3} \cdot \int \frac{3x+7-7}{3x+7} \hspace{0.1cm} dx = $$

$$= \frac{2}{3} \cdot \int \left( \frac{3x+7}{3x+7} - \frac{7}{3x+7} \right) \hspace{0.1cm} dx = $$

$$= \frac{2}{3} \cdot \int \left( 1 - \frac{7}{3x+7} \right) \hspace{0.1cm} dx = $$

$$= \frac{2}{3} \cdot \left( \int 1 \hspace{0.1cm} dx - \int \frac{7}{3x+7} \hspace{0.1cm} dx \right) = $$

$$= \frac{2}{3} \cdot \left( \int 1 \hspace{0.1cm} dx - 7 \cdot \int \frac{1}{3x+7} \hspace{0.1cm} dx \right) = $$

$$= \frac{2}{3} \cdot \left( \int 1 \hspace{0.1cm} dx - \frac{7}{3} \cdot \int \frac{3}{3x+7} \hspace{0.1cm} dx \right) = $$

$$= \frac{2}{3} \cdot \left( x - \frac{7}{3} \cdot \ln|3x+7| \right) = $$

$$= \frac{2x}{3} - \frac{14}{9} \cdot \ln|3x+7| + C $$

The general rule for trigonometric substitution is based on the expressions we have in the integral. If the integral contains:

- \(\sqrt{a^2-x^2}\) then \(x=a \cdot \sin(u)\). The very first example to practice this method is \(\int \frac{1}{\sqrt{1-x^2}} \hspace{0.1cm} dx\). Click here to watch the video.
- \(\sqrt{a^2+x^2}\) then \(x=a \cdot \tan(u)\). The very first example to practice this method is \(\int \frac{1}{\sqrt{1+x^2}} \hspace{0.1cm} dx\). Click here to watch the video.
- \(\sqrt{x^2-a^2}\) then \(x= \frac{a}{\cos(u)}\). The very first example to practice this method is \(\int \frac{1}{\sqrt{x^2-1}} \hspace{0.1cm} dx\). Click here to watch the video.

You can find lots of examples of trigonometric substitution in the Integration by trigonometric substitution section.