Let $f$ be any function and $F$ its antiderivative. The set of all antiderivatives of $f$ is called **indefinite** **integral** of the function $f$ denoted by

$$\int f(x) dx = F(x) + C.$$

Every two primitive functions differ by a constant $C$. This means, if we know the one antiderivative $F$ of the function $f$, then the another we can write in the form $F(x) + C$.

We do not have strictly rules for calculating the antiderivative (indefinite integral). The most antiderivatives we know is derived from the table of derivatives, which we read in the opposite direction.

**Table of basic integrals**

$$\int dx = x + C$$

$$\int x^n dx = \frac{x^{n+1}}{n+1} + C, \quad n\neq 1$$

$$\int \frac{1}{x} dx = \ln |x| + C$$

$$\int \frac{1}{x^n} dx = – \frac{1}{(n-1)x^{n-1}} + C, \quad n \neq 1$$

$$\int \frac{1}{\sqrt{x}} dx = 2 \sqrt{x} + C$$

$$\int e^x dx = e^x + C$$

$$\int a^x dx = \frac{a^x}{\ln a} + C, \quad a>0, a\neq 1$$

$$ \int \sin x dx = – \cos x + C$$

$$\int \cos x dx = \sin x + C$$

**Table of integrals of rational functions**

**Example 1**. Determine

$$\int (2 – 3x + x^2 )dx.$$

**Solution**.

By the additivity and linearity property of the integral, we have

$$\int (2 – 3x + x^2 )dx = 2 \int dx – 3 \int x dx + \int x^2 dx. $$

From the table of integral, we read

$$\int (2 – 3x + x^2 )dx = 2x – \frac{3}{2} x^2 + \frac{1}{3}x^3 + C.$$

**Example 2**. Determine

$$ \int \frac{x^3 -1}{2x^2} dx.$$

**Solution**.

$$\int \frac{x^3 -1}{2x^2} dx = \int \left(\frac{x^3}{2x^2} – \frac{1}{2x^2} \right) dx $$

$$= \int \left(\frac{x}{2} – \frac{1}{2x^2} \right) dx$$

$$ = \frac{1}{2} \left( \int x – \frac{1}{x^2} \right) dx.$$

By the additivity property of the integral, we have

$$\frac{1}{2} \left( \int x – \frac{1}{x^2} \right) dx = \frac{1}{2} \left (\int x dx – \int \frac{1}{x^2} dx \right). $$

From the table of indefinite integrals, we read:

$$\frac{1}{2} \left (\int x dx – \int \frac{1}{x^2} dx \right) = \frac{1}{2} \left ( \frac{1}{2}x^2 + \frac{1}{x} \right) + C $$

$$ = \frac{x^2}{4} + \frac{1}{2x} + C,$$

that is

$$\int \frac{x^3 -1}{2x^2} dx = \frac{x^2}{4} + \frac{1}{2x} + C.$$

**Example 3**. Determine

$$\int \frac{\sin 2x + 3 \sin^2 x}{\sin x } dx .$$

**Solution**.

Since $\sin 2x = 2 \sin x \cos x$, we can write

$$\int \frac{\sin 2x + 3 \sin^2 x}{\sin x } dx = \int \frac{2 \sin x \cos x + 3 \sin^2 x}{\sin x} dx $$

$$ = \int \left( \frac{2 \sin x \cos x}{\sin x} + \frac{3 \sin^2 x}{\sin x} \right) dx$$

$$= \int \left( 2 \cos x + 3 \sin x \right) dx.$$

By the linearity and additivity property of the integral, we have

$$\int \left( 2 \cos x + 3 \sin x \right) = 2 \int \cos x dx + 3 \int \sin x dx.$$

From the table of indefinite integrals we read:

$$ 2 \int \cos x dx + 3 \int \sin x dx = 2 \sin x – 3 \cos x + C,$$

that is

$$\int \frac{\sin 2x + 3 \sin^2 x}{\sin x } dx = 2 \sin x – 3 \cos x + C.$$

**Integration by substitution**

We will use the reverse chain rule of differentiation of composite functions and thus obtain the method which is called the method of substitution.

The chain rule:

$$\frac{d}{dx}f(g(x)) = f'(g(x)) \cdot g'(x).$$

The reverse of the chain rule:

$$\int f’ (g(x)) g'(x) dx = f(g(x)) + C.$$

**The integration rule**: if $g: I \to \mathbb{R}, I \subseteq \mathbb{R}$ is a differentiable and continuous function on $I$, then

$$\int f(g(x)) g'(x) dx = \int f(u) du,$$

where $g(x) = u$ and $g'(x) dx = du$.

Firstly, we need to choose a substitution to make. A substitution is not determined in advance, just by using the exercises we can discover the simplest way. Then we have to express the variable $x$ by $u$ and connect $du$ and $dx$. We change the integrand and $dx$ with expressions by the variable $u$ and calculate the integral. The result we write as the function of the variable $x$, that is, we return a substitution.

**Example 4**. Determine

$$ \int \sqrt{3 + x} dx.$$

**Solution**.

We choose a substitution $u= \sqrt{3+x}, (u \ge 0).$ It follows

$$x= u^2 – 3$$

and

$$dx = 2u du.$$

Now, by changing $x$ with $u$, we have:

$$\int \sqrt{3 + x} dx = \int u 2u du = \int 2u^2 du.$$

From the linearity property of integrals we can write

$$ \int 2u^2 du = 2 \int u^2 du.$$

From the table of integrals we read:

$$2 \int u^2 du = \frac{2}{3} u^3 + C.$$

Finally, the result we need to write as the function of the variable $x$:

$$\frac{2}{3} u^3 + C = \frac{2}{3} \sqrt{(3+x)^3} + C,$$

that is

$$\int \sqrt{3 +x } dx = \frac{2}{3} \sqrt{(3+x)^3} + C.$$

**Example 5**. Determine

$$\int \frac{2x + 1}{x^2 + x -3} dx.$$

**Solution**.

We choose a substitution $u = x^2 + x – 3$. Therefore, we have

$$\frac{du}{dx} = 2x +1 \Rightarrow dx = \frac{du}{2x+1}.$$

By replacing $x^2 + x – 3$ with $u$ and $dx$ with $\frac{du}{2x+1}$ we obtain:

$$\int \frac{2x + 1}{x^2 + x -3} dx = \int \frac{2x+1}{u} \cdot \frac{du}{2x + 1} = \int \frac{du}{u}.$$

From the table of integrals we read:

$$\int \frac{du}{u} = \ln |u|+ C.$$

By returning a substitution, we obtain the finally result

$$ \int \frac{2x + 1}{x^2 + x -3} dx = \ln |x^2 + x – 3| + C.$$

**Example 6**. Determine

$$\int \frac{\ln x}{x} dx.$$

**Solution**.

We choose a substitution $u = \ln x$. It follows

$$\frac{du}{dx} = \frac{1}{x} \Rightarrow dx = x du.$$

By replacing $\ln x$ with $u$ and $dx$ with $x du$, we have

$$\int \frac{\ln x}{x} dx = \int \frac{u}{x} x du = \int u du.$$

From the table of integrals we read:

$$\int u du = \frac{1}{2} u^2 + C,$$

that is

$$\int \frac{\ln}{x} dx = \frac{1}{2} \ln^2 x + C.$$

**Example 7**. Determine

$$\int \cot x dx.$$

**Solution**.

Since $\cot x = \frac{\cos x}{\sin x}$, we can write

$$\int \cot x dx = \int \frac{\cos x}{\sin x} dx.$$

We choose a substitution $u = \sin x$. It follows

$$\frac{du}{dx} = \cos x \Rightarrow dx = \frac{du}{\cos x}.$$

Therefore, we have

$$\int \frac{\cos x}{\sin x} dx = \int \frac{\cos x}{u} \cdot \frac{du}{\cos x} = \int \frac{du}{u}.$$

From the table of integrals we read

$$\int \frac{du}{u} = \ln|u| + C.$$

By returning a substitution, we obtain

$$\int \cot x dx = \ln |\sin x| + C.$$

The same principle we use for calculating the definite integral. However, the result we do not need write as the function of the variable $x$, because the result is a number. Instead, together with the integrand we are changing the lower and upper limit of integration.

**Example 8**. Determine

$$\int_{0}^{\frac{\pi}{2}} \cos^4 x \sin^3 x dx.$$

**Solution**.

We choose $u = \cos x$. It follows

$$\frac{du}{dx} = – \sin x \Longrightarrow dx = – \frac{du}{\sin x}.$$

The lower and upper limit are transformed in the following way:

$$x = 0 \Longrightarrow u = \cos 0 = 1,$$

$$x = \frac{\pi}{2} \Longrightarrow u = \cos \frac{\pi}{2} = 0.$$

Now we have:

$$\int_{0}^{\frac{\pi}{2}} \cos^4 x \sin^3 x dx = \int_{0}^{1}u^4 \cdot \sin^3 x \cdot \left( – \frac{du}{\sin x} \right) $$

$$= \int_{0}^{1} u^4 \cdot (-\sin^2) du.$$

From the relation $\sin^2 x + \cos ^2 x = 1$, we can write $\sin^2 x = 1 – \cos^2 x$, that is

$$ \int_{0}^{1} u^4 \cdot (-\sin^2) du = \int_{0}^{1} u^4 \cdot (-1)(1 – \cos^2 x) du $$

$$= \int_{0}^{1} u^4 \cdot (u^2 -1) du$$

$$ = \int_{0}^{1} (u^6 – u^4) du$$

$$= \int_{0}^{1} u^6 du – \int_{0}^{1} u^4 du.$$

By using the first fundamental theorem of calculus, finally we have:

$$\int_{0}^{1} u^6 du – \int_{0}^{1} u^4 du = \frac{u^7}{7} \bigg|_{0}^{1} – \frac{u^5}{5}\bigg|_{0}^{1} = – \frac{2}{35}$$

**Integration by parts**

We cannot calculate all integrals by using the method of substitution. Therefore, for calculate the simple integral as $\int x \ln x$ we need to use different methods.

Integration by parts is the one useful method for calculating integrals. Let $u$ and $v$ be functions of the variable $x$. The formula for derivative of the product

$$[u(x)v(x)]’ = u'(x)v(x) + u(x)v'(x)$$

can be written in the equivalent form:

$$u(x)v(x) = \int [u'(x)v(x) + u(x)v'(x)] dx = \int u'(x) v(x) dx + \int u(x)v'(x) dx.$$

Thus we obtain:

$$\int u(x)v'(x) dx = u(x)v(x) – \int u'(x)v(x) dx, $$

or more simply

$$\int u v’ dx = uv – \int u’ v dx.$$

The formula above is called the** formula for integration by parts**.

**The method of the integration by parts**

- The integrand write as the product of functions $u$ and $v’$.
- Calculate the extra integral $v= \int v’ dx$. The function $v’$ must be chosen in the way that its integral is easy to determine.
- Calculate the derivative $u’$.
- Write the formula for integration by parts:

$$\int uv’ dx = uv – \int u’v dx.$$

- Calculate the integral $\int u’v dx$.

**Example 9**. Determine

$$\int x \sin x dx.$$

**Solution**.

We choose $v’ = \sin x dx$ and $u = x$. Now we calculate the integral $v= \int v’ dx$, that is

$$v’ = \sin x \Rightarrow v= \int \sin x dx = – \cos x + C.$$

The derivative of $u$ is equal to

$$u = x \Rightarrow u’ = 1.$$

By using the formula for integration by parts we have

$$\int x \sin x dx =x \cdot( – \cos x) + \int \cos x.$$

From the table of integrals we read: $\int \cos x = \sin x + C$. Finally, we have

$$\int x \sin x dx = – x \cdot \cos x + \sin x + C.$$

In some cases, integration by parts must be repeated.

**Example 10**. Determine

$$\int e^x \cos x dx.$$

**Solution**.

We choose $u = e^x$ and $v’ = \cos x dx$. It follows

$$v = \int \cos x dx \Longrightarrow v = \sin x ,$$

$$du = e^x dx.$$

By using the formula for integration by parts, we have:

$$\int e^x \cos x dx =e^x \cdot \sin x – \int e^x \sin x dx.$$

Now we choose $t = e^x$ and $w’ = \sin x dx$. It follows

$$w = \int \sin x dx \Longrightarrow w = – \cos x ,$$

$$dt = e^x dx,$$

that is

$$\int e^x \cos x dx =e^x \cdot \sin x – \int e^x \sin x dx$$

$$= e^x \cdot \sin x – \left[e^x \cdot ( – \cos x) – \int e^x \cos x dx \right]$$

$$ = e^x \cdot \sin x + e^x \cdot \cos x – \int e^x \cos x dx.$$

Finally, we have

$$2 \int e^x \cos x dx = e^x ( \sin x + \cos x) + C,$$

that is

$$\int e^x \cos x dx = \frac{1}{2} e^x ( \sin x + \cos x) + C.$$

For the definite integral is valid the formula for integration by parts:

$$\int_{a}^b u(x) v'(x) dx = u(x) v(x) \bigg|_{a}^{b} – \int_{a}^{b} u'(x) v(x) dx.$$

**Example 11**. Determine

$$\int_{2}^{3} x e ^{x} dx.$$

**Solution**.

We choose $u= x$ and $v’= e^x dx$. It follows

$$v = \int e^x dx \Longrightarrow v = e^x,$$

$$du =dx.$$

By using the formula for integration by parts for definite integral and first fundamental theorem of calculus, we have:

$$\int_{2}^{3} x e ^{x} dx = x \cdot e^x \bigg|_{2}^3 – \int_{2}^{3} e^x dx $$

$$= x \cdot e^x \bigg|_{2}^3 – e^x \bigg|_2^3 $$

$$=(3 \cdot e^3 – 2 \cdot e^2) – (e^3 – e^2) $$

$$=2 e^3 – e^2 $$

$$=e^2(2e -1).$$