# Calculus

## Integration of a Logarithmic Form

The power rule for integration is valid for all values except when the exponent is equal to $$-1$$. This is because $\frac{d(\ln u)}{dx} = \frac{1}{u}\frac{du}{dx}$

Thus, reversing the process where the denominator's exponent is $$-1$$ would lead to an integral of the logarithmic form.

 $\int\frac{1}{u}du=\ln |u|+C$

Thus, the integration of $\int \frac{x \,dx}{3+x^2}$ will be done using the logarithmic form, whereas $\int \frac{x\, dx}{(3+x^2)^2}$ will be done using the power rule for integration.

Example 1: Integrate the function $\int_{1}^{2}\frac{1}{8-3x}dx$

Solution: We can recognize this is an integral of logarithmic form because the denominator is to the power of -1 (e.g., it can be written as $$(8-3x)^{-1}$$.

Let $$u=8-3x$$, $$du=-3dx$$. We can substitute these values and change the variable to u

$\frac{1}{-3}\int_{1}^{2}\frac{1}{u}du=\frac{1}{-3}\left[\ln |u|\right]_{1}^{2}$

Solving for the definite integral we get,

$\frac{\ln 2}{-3}-\frac{\ln 1}{-3} = \frac{\ln 2}{-3}\qquad \qquad (\ln 1=0)$

Example 2: Integrate the function $\int\frac{dx}{x(1+2\ln x)}$

Solution: Let $$u=1+2\ln x$$

\begin{align} du &= \frac{2}{x}dx \\ \frac{dx}{x} &= \frac{du}{2} \end{align}

The integral after subitution is

\begin{align} &\int \frac{du}{2u} \\ &=\frac{\ln |u|}{2} +C \\ &=\frac{\ln |1+2\ln x|}{2} +C \end{align}

Example 3: Integrate $\int \sec x\,dx$

Solution: There is a trick here and that is to multiply $$\sec x$$ by $\frac{\sec x + \tan x}{\sec x + \tan x}$

The resulting integral is

$\int \frac{\sec ^2x + \sec x\tan x}{\sec x + \tan x}dx$

Now let $$u=\sec x + \tan x$$

$du=\sec x\tan x + \sec ^2x \,dx$

The integral becomes,

$\int \frac{du}{u} = \ln |u| + C = \ln |\sec x + \tan x| + C$