Integration of Power and Logarithmic Functions

Power Integration

Integration of a power function, x to the power of any number, appears on our formula sheet as
$\int{x^{n}}dx=\frac{1}{n+1}x^{n+1}+C, n \neq -1$

There are a few important things to take note in this formula.

First every time we perform an integration we have to add in an arbitrary constant. This is because whenever we find the derivative of a function we eliminate any terms that are not a coefficient of x. Any terms that are just a sole number. Therefore when we integrate we do not regain this number so have to add in an arbitrary letter to represent this number. If we are given the value of any point on this function, we will be able to calculate C.

Second if we ever want to check our integration was correct we just need to differentiate it and we will get our original derivative. The C value will disappear.

Thirdly if n is equal to -1 then this formula will not work. This makes sense since we can never differentiate a power function to equal $x^{-1}$ and if we tried to integrate it we would end up having to divide the function by 0.

Instead for the specific case of $x^{-1}$ we are given the formula of:

$\int{\frac{1}{x}}dx=log_{e}|x|+C$

Power Integration Example

2013 exam question 2 a

$\int{(4-2x)^{-5}}dx$

We can use the power anti differentiation formula, however there is also a function inside a function, so we need to use a reverse chain rule.

Step 1 Choose our u value

$u=4-2x$

Step 2 Now reform our original equation:

$\int{u^{-5}}dx \times \frac{du}{dx} \times \frac{dx}{du} \newline \int{u^{-5}}du \times \frac{dx}{du}$

note: all we have done is multiply the original equations by fractions which would cancel out to one.

Step 3 Solve $\int{u^{-5}}du$ using our power equation

$\int{u^{-5}}du=-\frac{1}{4}u^{4}$

Step 4 Find $latex \frac{dx}{du}$

$u=4-2x \newline \frac{du}{dx}=-2 \newline \frac{dx}{du}=-\frac{1}{2}$

Step 5 Multiply the two parts from step 3 and 4 together

$\int{u^{-5}}du=-\frac{1}{4}u^{-4} \times -\frac{1}{2} \newline \newline =\frac{1}{8}u^{4}+C$

Step 6 Then substitute in the value of u

$\int{(4-2x)^{-5}}du =\frac{1}{8}(4-2x)^{4}+C$

At this stage if we are not confident of our answer we can always use the chain rule to differentiate the function and check our answer.

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