8.1. Integrals involving $f'(x)$

Illustrative examples:

$\displaystyle \int \frac{x}{x^2-5} \; \mathrm{d}x = \frac{\ln |x^2-5|}{2} + C$

$\displaystyle \int \frac{x}{\sqrt{x^2-5}} \; \mathrm{d}x = \sqrt{x^2-5} + C$

$\displaystyle \int x \, \mathrm{e}^{x^2-5} \; \mathrm{d}x = \frac{\mathrm{e}^{x^2-5}}{2} + C$

The formulas

The formulas below were used in solving the illustrative examples (arbitrary constants omitted):

$\displaystyle \int \frac{f'(x)}{f(x)} \; \mathrm{d}x = \ln | f(x) | \\ \displaystyle \int f'(x) \big ( f(x) \big )^n \; \mathrm{d}x = \frac{ \big (f(x) \big )^{n+1}}{n+1} \qquad \mathrm{for} \; n \neq -1 \\ \displaystyle \int f'(x) \; \mathrm{e}^{f(x)} \; \mathrm{d}x = \mathrm{e}^{f(x)}$

For each of the 3 examples, we let $f(x) = x^2-5$ and have $f'(x) = 2x$.

Make sure we understand when to use the 1st vs the 2nd formula. In particular, for all powers except $-1$, the second formula is used. Thus we let $n=-\frac{1}{2}$ for the second example.

Also note that these formulas will not be provided during examinations.

Understanding the formulas: integration as the reverse of differentiation

Let us differentiate the following: $\displaystyle \frac{\mathrm{d}}{\mathrm{d}x} \ln ( x - \tan x) = \frac{1}{x- \tan x} \cdot (1 - \mathrm{sec}^2 x) \\ \displaystyle \frac{\mathrm{d}}{\mathrm{d}x} ( x^2 + x + 2)^{\frac{1}{2}} = \frac{1}{2} ( x^2 + x + 2)^{-\frac{1}{2}} \cdot (2x+1) \\ \displaystyle \frac{\mathrm{d}}{\mathrm{d}x} \mathrm{e}^{\sin x} = \mathrm{e}^{\sin x} \cdot \cos x$.

We observe that, to differentiate $ \ln ( x - \tan x), ( x^2 + 2x + 2)^{\frac{1}{2}}$ and $\mathrm{e}^{\sin x}$, we first apply the formulas for differentiating $\ln X, X^{\frac{1}{2}}$ and $\mathrm{e}^X$ respectively. We then differentiate $(x - \tan x), (x^2 + 2x + 2)$ and $\sin x$ respectively. Chain rule then says to multiply the results from the first and second parts. Put abstractly, $\displaystyle \frac{\mathrm{d}}{\mathrm{d}x} \ln f(x) = \frac{f'(x)}{f(x)}, \\ \displaystyle \frac{\mathrm{d}}{\mathrm{d}x} \big (f(x) \big )^n = f'(x) \Big (n \big (f(x) \big )^{n-1} \Big) \\ \displaystyle \frac{\mathrm{d}}{\mathrm{d}x} \mathrm{e}^{f(x)} = f'(x) \mathrm{e}^x.$

We understand integration as the reverse of differentiation. Thus, the three examples teaches how to perform the following 3 integrals: $\displaystyle \int \frac{1-\mathrm{sec}^2 x}{x - \tan x} \; \mathrm{d} x = \ln |x - \tan x| + C \\ \displaystyle \int \frac{2x+1}{2} ( x^2 + x + 2)^{-\frac{1}{2}} \; \mathrm{d}x = ( x^2 + 2x + 2)^{\frac{1}{2}} + C \\ \displaystyle \int \cos x \, \mathrm{e}^{\sin x} \; \mathrm{d}x = \mathrm{e}^{\sin x} + C$.

In other words, while we may not know how to integrate $\frac{1}{x - \tan x}$ , we now know how to integrate $\frac{1-\mathrm{sec}^2 x}{x - \tan x}$. Similarly, we do not have a method to integrate $( x^2 + x + 2)^{-\frac{1}{2}}$ and $\mathrm{e}^{\sin x}$, but we do know how to integrate $(2x+1) ( x^2 + x + 2)^{-\frac{1}{2}}$ and $\cos x \, \mathrm{e}^{\sin x}$. Let $f(x)$ denote $(x-\tan x), (x^2+ x+2)$ and $\sin x$ for each of the examples respectively. The common feature behind the 3 new integrals we have discovered is the presence of the term $f'(x)$ in our integrand.

Put abstractly, we arrive at our 3 $f'(x)$ formulas: $\displaystyle \int \frac{f'(x)}{f(x)} \; \mathrm{d}x = \ln | f(x) | + C \\ \displaystyle \int f'(x) \big ( f(x) \big )^n \; \mathrm{d}x = \frac{ \big (f(x) \big )^{n+1}}{n+1} + C \qquad \mathrm{for} \; n \neq -1 \\ \displaystyle \int f'(x) \; \mathrm{e}^{f(x)} \; \mathrm{d}x = \mathrm{e}^{f(x)}+ C$.

Obtaining $f'(x)$: multiplying by $\displaystyle \frac{a}{a}$

For the example questions $\displaystyle \int \frac{x}{x^2-5} \mathrm{d}x$, $\displaystyle \int \frac{x}{\sqrt{x^2-5}} \; \mathrm{d}x$ and $\displaystyle \int x \, \mathrm{e}^{x^2-5} \; \mathrm{d} x$, we let $f(x)$ denote $x^2-5$. Usage of the formulas require $f'(x) = 2x$. However, the integrand for each of the examples only contain $x$: it is missing a 2.

Constants such as $2$ can be obtained by the following trick. Notice that multiplication by 1 does not change an expression. We thus multiply our expressions by $\frac{2}{2}$ to get, for the first example, $\displaystyle \int \frac{x}{x^2-5} \mathrm{d}x = \int \frac{2x}{2 (x^2-5)} \; \mathrm{d}x = \frac{1}{2} \int \frac{2x}{x^2-5} \; \mathrm{d}x$. We thus have "forced" out $f'(x)$ to use the formulas discussed above.

Similarly, $\displaystyle \int \frac{x}{\sqrt{x^2-5}} \; \mathrm{d}x = \frac{1}{2} \int (2x) (x^2-5)^{-\frac{1}{2}} \; \mathrm{d}x = \frac{1}{2} \frac{ (x^2-5)^{\frac{1}{2} } }{ \frac{1}{2} } + C$. The same trick is also applicable for the third example.

A additional exercise: find $\displaystyle \int \frac{\ln x}{x} \; \mathrm{d}x$ $\displaystyle \frac{ (\ln x)^2 }{2} + C$
Hover over or click on the integral for the answer.