Worked examples · Sheet 1

Basic integrals and integrals in composite form

Theory used: Indefinite integral Basic integrals Basic integrals in composite form Antiderivatives of the same function

Go to: Exercise 1 Exercise 2 Exercise 3 Exercise 4

Exercise 1

Compute

\int (6 x^{2} - 4 x + 3 - \frac{2}{x}) d x .

\int (6 x^{2} - 4 x + 3 - \frac{2}{x}) d x = 6 \int x^{2} d x - 4 \int x d x + 3 \int d x - 2 \int \frac{1}{x} d x = 6 \cdot \frac{x ^{3}}{3} - 4 \cdot \frac{x ^{2}}{2} + 3 x - 2 ln ∣ x ∣ + c = 2 x^{3} - 2 x^{2} + 3 x - 2 ln ∣ x ∣ + c

By the linearity of the indefinite integral, each term is integrated separately, bringing the multiplicative constants outside. The powers $x^{2}$ and $x$ are integrated with the formula $\frac{x ^{a + 1}}{a + 1}$ , the constant $3$ has antiderivative $3 x$ , and the term $\frac{1}{x}$ , which is the only power excluded from that formula, has antiderivative $ln ∣ x ∣$ . The constants of the individual integrals are combined into a single constant $c$ . As a check, the derivative of the result is $6 x^{2} - 4 x + 3 - \frac{2}{x}$ , that is, the integrand.

Exercise 2

Compute, for $x > 0$ ,

\int \frac{( x + 1 ) ^{2}}{x} d x .

\frac{( x + 1 ) ^{2}}{x} = \frac{x ^{2} + 2 x + 1}{x ^{\frac{1}{2}}} = x^{\frac{3}{2}} + 2 x^{\frac{1}{2}} + x^{- \frac{1}{2}}

\int (x^{\frac{3}{2}} + 2 x^{\frac{1}{2}} + x^{- \frac{1}{2}}) d x = \frac{x ^{\frac{5}{2}}}{\frac{5}{2}} + 2 \cdot \frac{x ^{\frac{3}{2}}}{\frac{3}{2}} + \frac{x ^{\frac{1}{2}}}{\frac{1}{2}} + c = \frac{2}{5} x^{\frac{5}{2}} + \frac{4}{3} x^{\frac{3}{2}} + 2 x + c

The quotient does not have a basic antiderivative, so it is first transformed into a sum of powers. We expand the square of the binomial and divide each term by $x = x^{\frac{1}{2}}$ , subtracting $\frac{1}{2}$ from the exponents. Each power has exponent different from $- 1$ , therefore it is integrated with the power formula, and dividing by $\frac{5}{2}$ , $\frac{3}{2}$ , $\frac{1}{2}$ is equivalent to multiplying by $\frac{2}{5}$ , $\frac{2}{3}$ , $2$ . The last term is rewritten as $2 x$ .

Exercise 3

Compute

(a) \int \frac{2 x + 1}{x ^{2} + x + 5} d x, (b) \int x (x^{2} - 4)^{5} d x .

(a) (x^{2} + x + 5)^{'} = 2 x + 1, \int \frac{2 x + 1}{x ^{2} + x + 5} d x = ln (x^{2} + x + 5) + c

(b) \int x (x^{2} - 4)^{5} d x = \frac{1}{2} \int 2 x (x^{2} - 4)^{5} d x = \frac{1}{2} \cdot \frac{( x ^{2} - 4 ) ^{6}}{6} + c = \frac{( x ^{2} - 4 ) ^{6}}{12} + c

In the first integral the numerator is exactly the derivative of the denominator, so the integral has the composite form $\int \frac{g ^{'} ( x )}{g ( x )} d x = ln ∣ g (x)∣ + c$ with $g (x) = x^{2} + x + 5$ . The discriminant of $g$ is $1 - 20 < 0$ and the coefficient of $x^{2}$ is positive, therefore $g (x) > 0$ for every $x$ and the absolute value can be omitted.

In the second integral the power $g (x)^{5}$ appears with $g (x) = x^{2} - 4$ , whose derivative is $2 x$ . The factor $x$ present in the integrand is the derivative up to the factor $2$ , so we multiply and divide by $2$ and apply the composite form $\int g^{a} g^{'} d x = \frac{g ^{a + 1}}{a + 1} + c$ with $a = 5$ . Differentiating the result gives $\frac{6 ( x ^{2} - 4 ) ^{5} \cdot 2 x}{12} = x (x^{2} - 4)^{5}$ , as it must be.

Exercise 4

Find the antiderivative $F$ of

f (x) = \frac{6 x}{3 x ^{2} + 1} + e^{2 x}

such that $F (0) = 3$ .

\int (\frac{6 x}{3 x ^{2} + 1} + e^{2 x}) d x = ln (3 x^{2} + 1) + \frac{e ^{2 x}}{2} + c

F (0) = ln 1 + \frac{1}{2} + c = \frac{1}{2} + c = 3 ⟺ c = \frac{5}{2}

F (x) = ln (3 x^{2} + 1) + \frac{e ^{2 x}}{2} + \frac{5}{2}

The derivative of $3 x^{2} + 1$ is $6 x$ , so the first term has the composite form $\frac{g ^{'}}{g}$ and has antiderivative $ln (3 x^{2} + 1)$ , without absolute value because $3 x^{2} + 1 > 0$ . The second term has the form $e^{g (x)} g^{'} (x)$ with $g (x) = 2 x$ up to the factor $2$ , therefore its antiderivative is $\frac{e ^{2 x}}{2}$ . By the theorem on antiderivatives of the same function, all the antiderivatives of $f$ are obtained by adding a constant $c$ , and the condition $F (0) = 3$ singles out only one of them. Since $ln 1 = 0$ and $e^{0} = 1$ , the condition becomes $\frac{1}{2} + c = 3$ , so $c = \frac{5}{2}$ .