5.9 Indefinite Integral

1. The Indefinite Integral

Consider $F(x) = x^2$. The derivative is $F'(x) = 2x$.

The reverse problem: If they give us the result $f(x) = 2x$, can we find a function $F(x)$, such that $F'(x) = f(x)$?

Of course, one answer is $F(x) = x^2$. We say that $F(x) = x^2$ is an antiderivative of $f(x) = 2x$.

But it is not the only one! Notice that:

$(x^2)' = 2x$
$(x^2 + 1)' = 2x$
$(x^2 + 2)' = 2x$
$(x^2 + 5)' = 2x$

In general, $(x^2 + c)' = 2x$ for any constant $c$. Therefore, the functions $x^2 + c$ are also antiderivatives of $f(x) = 2x$.

We say that $x^2 + c$ is the indefinite integral of $f(x) = 2x$ and we use the notation:

$$\int 2x \, dx = x^2 + c$$

Hence, we establish the general rule:

If $F'(x) = f(x)$ then $\displaystyle\int f(x) \, dx = F(x) + c$

For example, since $(x^5)' = 5x^4$, we obtain $\int 5x^4 \, dx = x^5 + c$. We also deduce that $\int x^4 \, dx = \dfrac{1}{5}x^5 + c$. Therefore, we can easily obtain the following results for powers of $x$:

$f(x)$	$\int f(x) \, dx$
$1$	$x + c$
$x$	$\dfrac{x^2}{2} + c$
$x^2$	$\dfrac{x^3}{3} + c$
$x^3$	$\dfrac{x^4}{4} + c$
$x^{10}$	$\dfrac{x^{11}}{11} + c$

In general, the integral of the power function $f(x) = x^n$ is:

$$\int x^n \, dx = \dfrac{1}{n+1}x^{n+1} + c \quad \text{if } n \neq -1$$

Integrals of Most Common Functions

If we remember the derivatives of the basic functions we effortlessly obtain the following corresponding indefinite integrals. Notice also that $\int a \, dx = ax + c$ since $(ax)' = a$, and $\int \cos x \, dx = \sin x + c$ since $(\sin x)' = \cos x$.

$f(x)$	$\int f(x) \, dx$
$x^n$	$\dfrac{1}{n+1}x^{n+1} + c \quad (n \neq -1)$
$a$ (constant)	$ax + c$
$e^x$	$e^x + c$
$\sin x$	$-\cos x + c$
$\cos x$	$\sin x + c$
$\dfrac{1}{x}$	$\ln \|x\| + c$

NOTICE: The operational process of finding the integral is formally called integration.

2. Remarks on Negative and Rational Powers

The formula $\displaystyle\int x^n \, dx = \dfrac{1}{n+1}x^{n+1} + c$ applies perfectly for negative values of $n$. For example:

$$\int x^{-5} \, dx = \dfrac{x^{-4}}{-4} + c = -\dfrac{1}{4x^4} + c$$

What about $\displaystyle\int \dfrac{1}{x^2} \, dx$? We know that $\dfrac{1}{x^2} = x^{-2}$, so:

$$\int \dfrac{1}{x^2} \, dx = \int x^{-2} \, dx = \dfrac{x^{-1}}{-1} + c = -\dfrac{1}{x} + c$$

Also, the same formula applies for rational values of $n$

$$ \begin{aligned} \int x^{3/5} \, dx &= \dfrac{x^{8/5}}{8/5} + c = \dfrac{5}{8}x^{8/5} + c \\[0.2cm] \int \sqrt{x} \, dx &= \int x^{1/2} \, dx = \dfrac{x^{3/2}}{3/2} + c = \dfrac{2}{3}x^{3/2} + c \end{aligned} $$

$f(x)$	$\int f(x) \, dx$
$\dfrac{1}{x^2} = x^{-2}$	$\dfrac{x^{-1}}{-1} + c = -\dfrac{1}{x} + c$
$\dfrac{1}{x^3} = x^{-3}$	$\dfrac{x^{-2}}{-2} + c = -\dfrac{1}{2x^2} + c$
$\dfrac{1}{x^4} = x^{-4}$	$\dfrac{x^{-3}}{-3} + c = -\dfrac{1}{3x^3} + c$
$\sqrt[3]{x^2} = x^{2/3}$	$\dfrac{x^{5/3}}{5/3} + c = \dfrac{3}{5}x^{5/3} + c$

Remark for $\displaystyle\int \dfrac{1}{x} \, dx$ HL ONLY
Notice that the power formula does not apply for $\int \dfrac{1}{x} \, dx = \int x^{-1} \, dx$. Only for this particular power we have the strict formula $\int \dfrac{1}{x} \, dx = \ln|x| + c$.
Indeed:

If $x > 0$, then $[\ln|x|]' = [\ln x]' = \dfrac{1}{x}$
If $x < 0$, then $[\ln|x|]' = [\ln(-x)]' = \dfrac{1}{-x}(-1) = \dfrac{1}{x}$

3. Two Basic Rules of Integration

Rule (1): Addition and Subtraction

$$\int [f(x) + g(x)] \, dx = \int f(x) \, dx + \int g(x) \, dx$$

For example, $$ \int (x^4 + x^2) \, dx = \int x^4 \, dx + \int x^2 \, dx = \dfrac{x^5}{5} + \dfrac{x^3}{3} + c $$

Rule (2): Scalar Multiplication

$$\int a f(x) \, dx = a \int f(x) \, dx$$

For example, $$ \int 10x^4 \, dx = 10 \int x^4 \, dx = 10 \left(\dfrac{x^5}{5}\right) + c = 2x^5 + c $$

In practice, if we have a "long" expression like $$ \int [a f(x) + b g(x) + c h(x) + d k(x)] \, dx $$ where $a, b, c, d$ are constant coefficients, we keep $a, b, c, d$ and integrate only $f(x), g(x), h(x), k(x)$.

EXAMPLE 1

Solution:
$$ \begin{aligned} \int [3x^2 + 5e^x - 2\cos x] \, dx &= 3 \int x^2 \, dx + 5 \int e^x \, dx - 2 \int \cos x \, dx \\ &= 3\left(\dfrac{x^3}{3}\right) + 5e^x - 2\sin x + c \\ &= x^3 + 5e^x - 2\sin x + c \end{aligned} $$

Notice: The intermediate split step highlighted above is not strictly necessary in practice! We can proceed directly to the final result by evaluating the integrals of $x^2$, $e^x$, and $\cos x$ mentally.

EXAMPLE 2

Solution:
$$ \begin{aligned} \int [2x^4 + 8x^3 - 5x^2 - 7x + 2] \, dx &= \dfrac{2}{5}x^5 + \dfrac{8}{4}x^4 - \dfrac{5}{3}x^3 - \dfrac{7}{2}x^2 + 2x + c \\ &= \dfrac{2}{5}x^5 + 2x^4 - \dfrac{5}{3}x^3 - \dfrac{7}{2}x^2 + 2x + c \end{aligned} $$

EXAMPLE 3

Solution:
$$ \begin{aligned} \int \left(\dfrac{2}{x^4} + \dfrac{8}{x^3} - \dfrac{5}{x^2} + 2\right) \, dx &= \int [2x^{-4} + 8x^{-3} - 5x^{-2} + 2] \, dx \\ &= 2\left(\dfrac{x^{-3}}{-3}\right) + 8\left(\dfrac{x^{-2}}{-2}\right) - 5\left(\dfrac{x^{-1}}{-1}\right) + 2x + c \\ &= -\dfrac{2}{3x^3} - \dfrac{4}{x^2} + \dfrac{5}{x} + 2x + c \end{aligned} $$

4. Finding the Constant $c$

Sometimes, we are given an extra condition of the form $f(a) = b$ in order to evaluate the exact value of the constant integration coefficient $c$. We obtain, in fact, the specific unique antiderivative function that satisfies this precise geometric or numerical constraint.

EXAMPLE 4

Let $f'(x) = 6x^2 - 4x + 5$. Find $f(x)$ given that $f(1) = 8$.

Solution:
Clearly, $f(x)$ corresponds to the indefinite integral of the derivative function $f'(x) = 6x^2 - 4x + 5$. That is:
$$ \begin{aligned} f(x) &= \int [6x^2 - 4x + 5] \, dx \\ &= 6\left(\dfrac{x^3}{3}\right) - 4\left(\dfrac{x^2}{2}\right) + 5x + c \\ &= 2x^3 - 2x^2 + 5x + c \end{aligned} $$ Next, we must invoke the initial condition $f(1) = 8$ to evaluate the value of $c$:
$$ \begin{aligned} f(1) = 8 &\implies 2(1)^3 - 2(1)^2 + 5(1) + c = 8 \\ &\implies 5 + c = 8 \\ &\implies c = 3 \end{aligned} $$ Therefore, the explicit equation for the function is:

$f(x) = 2x^3 - 2x^2 + 5x + 3$