Worked examples · Sheet 2

Integrals by substitution

Theory used: Integration by substitution Substitution in definite integrals Basic integrals Fundamental formula of integral calculus

Go to: Exercise 1 Exercise 2 Exercise 3 Exercise 4

Exercise 1

Compute

\int x \sqrt{x + 1}\,dx .

t = x + 1, \qquad x = t - 1, \qquad dx = dt

\begin{aligned} \int x \sqrt{x + 1}\,dx &= \int (t - 1)\,t^{\frac{1}{2}}\,dt = \int \left( t^{\frac{3}{2}} - t^{\frac{1}{2}} \right) dt \\[6pt] &= \frac{2}{5}\,t^{\frac{5}{2}} - \frac{2}{3}\,t^{\frac{3}{2}} + c \\[6pt] &= \frac{2}{5}\,(x + 1)^{\frac{5}{2}} - \frac{2}{3}\,(x + 1)^{\frac{3}{2}} + c \end{aligned}

The root contains $x + 1$ , while outside the root there is $x$ ; setting $t = x + 1$ , both factors become powers of $t$ . From the substitution follows $x = t - 1$ and, differentiating, $dx = dt$ . Multiplying $(t - 1)$ by $t^{\frac{1}{2}}$ gives a difference of powers, which is integrated with the power formula. At the end we return to the variable $x$ by substituting $t = x + 1$ .

Exercise 2

Compute, for $x > 0$ ,

\int \frac{1}{x + \sqrt{x}}\,dx .

t = \sqrt{x}, \qquad x = t^2, \qquad dx = 2t\,dt

\int \frac{1}{x + \sqrt{x}}\,dx = \int \frac{2t}{t^2 + t}\,dt = \int \frac{2t}{t\,(t + 1)}\,dt = \int \frac{2}{t + 1}\,dt = 2 \ln(t + 1) + c

\int \frac{1}{x + \sqrt{x}}\,dx = 2 \ln\left( 1 + \sqrt{x} \right) + c

The substitution $t = \sqrt{x}$ eliminates the root, because $x = t^2$ and $\sqrt{x} = t$ for $x > 0$ and $t > 0$ . Differentiating $x = t^2$ with respect to $t$ gives $dx = 2t\,dt$ . The denominator becomes $t^2 + t = t\,(t + 1)$ , and the factor $t$ cancels with the one in $2t$ . What remains is $\dfrac{2}{t + 1}$ , which has antiderivative $2 \ln \lvert t + 1 \rvert$ ; the absolute value is omitted because $t + 1 > 0$ . Finally we substitute $t = \sqrt{x}$ .

Exercise 3

Compute

\int \frac{e^{2x}}{e^{x} + 1}\,dx .

t = e^{x}, \qquad dt = e^{x}\,dx

\int \frac{e^{2x}}{e^{x} + 1}\,dx = \int \frac{e^{x}}{e^{x} + 1}\,e^{x}\,dx = \int \frac{t}{t + 1}\,dt

\int \frac{t}{t + 1}\,dt = \int \left( 1 - \frac{1}{t + 1} \right) dt = t - \ln(t + 1) + c

\int \frac{e^{2x}}{e^{x} + 1}\,dx = e^{x} - \ln\left( e^{x} + 1 \right) + c

We write $e^{2x} = e^{x} \cdot e^{x}$ . One of the two factors, together with $dx$ , forms the differential $dt = e^{x}\,dx$ of the new variable $t = e^{x}$ , and the other becomes $t$ . We obtain the rational function $\dfrac{t}{t + 1}$ , whose numerator and denominator have the same degree. Writing $t = (t + 1) - 1$ splits the quotient into $1 - \dfrac{1}{t + 1}$ , whose terms are basic integrals. Since $t + 1 = e^{x} + 1 > 0$ , the logarithm does not require the absolute value. Returning to the variable $x$ gives the result.

Exercise 4

Compute

\int_0^3 \frac{x}{\sqrt{x + 1}}\,dx .

t = \sqrt{x + 1}, \qquad x = t^2 - 1, \qquad dx = 2t\,dt, \qquad x = 0 \mapsto t = 1, \qquad x = 3 \mapsto t = 2

\begin{aligned} \int_0^3 \frac{x}{\sqrt{x + 1}}\,dx &= \int_1^2 \frac{t^2 - 1}{t}\,2t\,dt = 2 \int_1^2 \left( t^2 - 1 \right) dt \\[6pt] &= 2 \left[ \frac{t^3}{3} - t \right]_1^2 = 2 \left[ \left( \frac{8}{3} - 2 \right) - \left( \frac{1}{3} - 1 \right) \right] \\[6pt] &= 2 \left( \frac{2}{3} + \frac{2}{3} \right) = \frac{8}{3} \end{aligned}

We set $t = \sqrt{x + 1}$ to eliminate the root; squaring gives $x = t^2 - 1$ , and hence $dx = 2t\,dt$ . Since this is a definite integral, the bounds are changed as well; for $x = 0$ we have $t = \sqrt{1} = 1$ , for $x = 3$ we have $t = \sqrt{4} = 2$ . The factor $t$ in the denominator cancels with the one in $2t$ , and the polynomial $2\left( t^2 - 1 \right)$ remains. With the fundamental formula of calculus we compute the difference of the values of the antiderivative at the bounds $2$ and $1$ , without returning to the variable $x$ . The integral equals $\dfrac{8}{3}$ , and it is positive as it must be, because the integrand is positive on $(0, 3]$ .