Tedious Integration

Big Idea

We regard integration, at the pre-university level, as the reverse process of differentiation:

$\displaystyle F(x) = \int f(x)\, \mathrm dx \quad \iff \quad \frac{\mathrm d}{\mathrm dx} (F(x)) = f(x).$

We call $f(x)$ the integrand. Integration is linear in the following sense:

$\displaystyle \int (af(x) + bg(x))\, \mathrm dx = a \int f(x)\, \mathrm dx + b \int g(x)\, \mathrm dx.$

The integrals of powers requires a bit more care: if $n \neq -1$ then nothing weird happens:

$\displaystyle \int x^n\, \mathrm dx = \frac{x^{n+1}}{n+1} +C.$

However, if $n = -1$ , then we cannot divide by zero, and instead recover the logarithm:

$\displaystyle \int x^{-1}\, \mathrm dx = \int \frac 1x \, \mathrm dx = \ln|x| + C.$

Other commonly used integrals are free for use. Furthermore, in the context of pre-university mathematics, the definite integral is an application of the vanilla integral:

$\displaystyle \int_a^b f(x)\, \mathrm dx := F(b) - F(a) \equiv [F(x)]_a^b.$

Questions

You may not use integration by substitution in any of these problems.

Question 1. Evaluate $\displaystyle \int (3^{2x} + \sqrt{25x} - \ln(x^7))\, \mathrm dx$ .

(Click for Solution)

Solution. Simplifying the integrand then integrating term-wise,

$\begin{aligned} \int (3^{2x} + \sqrt{25x} - \ln(x^7))\, \mathrm dx &= \int (9^x + 5x^{1/2} - 7 \ln(x))\, \mathrm dx \\ &= \int 9^x\, \mathrm dx + 5 \int x^{1/2}\, \mathrm dx - 7 \int \ln(x)\, \mathrm dx \\ &= \frac{9^x}{\ln(9)} + 5 \cdot \frac {x^{3/2}}{3/2} - 7(x \ln(x) - x) + C \\ &= \frac {1}{\ln(9)} \cdot 9^x+ \frac{10}3 \cdot x^{3/2} - 7x \ln(x) + 7x + C. \end{aligned}$

Question 2. Evaluate $\displaystyle \int (2+3x)^4\, \mathrm dx$ .

(Click for Solution)

Solution. We first slowly expand the integrand:

$\begin{aligned} (2+3x)^4 &= ( (2+3x)^2 )^2 \\ &= ( 4 + 12x + 9x^2 )^2 \\ &= 16 + 96x + 216x^2 + 216x^3 + 81x^4. \end{aligned}$

Using linearity to integrate term-wise,

$\begin{aligned} \int (2+3x)^4 \, \mathrm dx &= \int (16 + 96x + 216x^2 + 216x^3 + 81x^4)\, \mathrm dx \\ &= \int 16\, \mathrm dx + 96 \int x^1\, \mathrm dx + 216 \int x^2\, \mathrm dx + 216 \int x^3\, \mathrm dx + 81 \int x^4\, \mathrm dx \\ &= 16x + 96 \cdot \frac{x^2}{2} + 216 \cdot \frac{x^3}{3} + 216 \cdot \frac{x^4}{4} + 81 \cdot \frac{x^5}{5} + C \\ &= 16x + 48x^2 + 72x^3 + 54 x^4 + \frac{81}{5} \cdot x^5 + C. \end{aligned}$

Question 3. Evaluate $\displaystyle \int \left( \frac 2x - 7 \sqrt x \right)^2\, \mathrm dx$ .

(Click for Solution)

Solution. We first slowly expand the integrand:

$\begin{aligned} \left( \frac 2x - 7 \sqrt x \right)^2 &= \left( \frac 2x \right)^2 - 2 \cdot \frac 2x \cdot 7\sqrt x + (7 \sqrt x)^2 \\ &= \frac 4{x^2} - 28x^{-1/2} + 49x \\ &= 4x^{-2}- 28x^{-1/2} + 49x^1. \end{aligned}$

Using linearity to integrate term-wise,

$\begin{aligned} \int \left( \frac 2x - 7 \sqrt x \right)^2\, \mathrm dx &= \int (4x^{-2}- 28x^{-1/2} + 49x^1)\, \mathrm dx \\ &= 4 \int x^{-2}\, \mathrm dx - 28 \int x^{-1/2}\, \mathrm dx + 49 \int x^1\, \mathrm dx \\ &= 4 \cdot \frac{x^{-1}}{-1} - 28 \cdot \frac{x^{1/2}}{1/2} + 49 \cdot \frac{x^2}{2} + C \\ &= - 4x^{-1}- 56 x^{1/2} + \frac{49}{2} \cdot x^2 + C. \end{aligned}$

Remark 1. In general, when feasible, final answers ought to take the form

$\displaystyle c_1 \cdot f_1(x) + c_2 \cdot f_2(x) + \cdots + c_n \cdot f_n(x) + C,$

where $c_1,\dots, c_n$ are numerical constants, $f_1(x),\dots, f_n(x)$ are expressions in terms of $x$ , and $C$ denotes the arbitrary constant of integration (if and only if needed), for easy recognition.