**(1) Basic indefinite integrals:**

$$\int x^a \ dx = \frac{x^{a+1}}{a+1} \ \ \ \ \ (a \ne -1) \tag{1}$$

$$\int \frac{1}{x}\ dx = \ln{\left|x\right|} \ \ \ \ \ \tag{2}$$

$$\int \sin{x}\ dx = -\cos{x} \ \ \ \ \ \tag{3}$$

$$\int \cos{x}\ dx = \sin{x} \ \ \ \ \ \tag{4}$$

$$\int \frac{1}{\cos^2{x}}\ dx = \tan{x} \tag{5}$$

$$\int \frac{1}{\sin^2{x}}\ dx = -\cot{x} \tag{6}$$

$$\int {\rm e}^x \ dx = {\rm e}^x \tag{7}$$

$$\int \frac{1}{1+x^2}\ dx = \arctan{x} \tag{8}$$

$$\int \frac{1}{\sqrt{1-x^2}}\ dx = \arcsin{x} \tag{9}$$

$$\int \frac{1}{\sqrt{x^2 + a}}\ dx = \ln {\left|x + \sqrt {x^2 + a}\right|} \ \ \ \ \ (a \ne 0) \tag{10}$$

Note that these formulas are valid for those values of $x$ for which the integrand function is defined. E.g. $(1)$, $(3)$, $(4)$ are valid for all $x$, $(2)$ is valid for non-zero values of $x$, $(5)$ is valid for values of $x$ such that $\cos{x} \ne 0$.

**(2) Additional indefinite integrals:**

These three integrals are quite important and often met. The formulas below can be derived via integration by parts.

$$\int \sqrt{a^2-x^2}\ dx = \frac{1}{2} ( x \sqrt {a^2-x^2} + a^2 \arcsin {\frac{x}{a}} ) \ \ \ \ \ \ (a \gt 0, \left|x\right| \lt a) \tag{1'}$$

$$\int \sqrt{x^2-a^2}\ dx = \frac{1}{2} ( x \sqrt {x^2-a^2} - a^2 \ln {\left|x + \sqrt {x^2-a^2}\right|} ) \ \ \ \ \ \ (a \gt 0, \left|x\right| \gt a) \tag{2'}$$

$$\int \sqrt{x^2+a^2}\ dx = \frac{1}{2} ( x \sqrt {x^2+a^2} + a^2 \ln {\left|x + \sqrt {x^2+a^2}\right|} ) \ \ \ \ \ \ (a \gt 0) \tag{3'}$$

**(3) Another additional indefinite integral**

Let us suppose we are given this integral which we want to calculate. This one is often met when trying to integrate rational functions.

$$ I(a, n) = \int \frac{dx}{(a^2+x^2)^n}$$

For this integral, one can prove via integration by parts the following recurrent relation:

$$ I(a, n) = \frac{1}{a^2}\cdot\frac{2n-3}{2n-2}\cdot I(a, n-1) + \frac{1}{(2n-2)a^2} \cdot \frac{x}{(a^2+x^2)^{n-1}}$$

Also, it is easy to see that:

$$I(a,1) = \frac{1}{a} \cdot \arctan{\frac{x}{a}}$$

The last two identities give us a procedure for calculating the integrals $I(a, n)$.