# Video: Integrate Exponential Functions

Find β«(π^(5 ln π₯) β π^(7π₯)) dπ₯.

02:54

### Video Transcript

Find the integral of π raised to the power of five multiplied by the natural logarithm of π₯ minus π raised to the power of seven π₯ with respect to π₯.

We start by noticing that our integrand is two terms. The second term we know how to integrate. However, the first term π raised to the power of five multiplied by the natural logarithm of π₯ is a composition of functions we donβt know how to integrate. So weβll need to manipulate this. To do this, weβre going to use the following two laws. The first tells us that π raised to the power of the natural logarithm of π is just equal to π. And the second one tells us that π multiplied by the natural logarithm of π is equal to the natural logarithm of π raised to the power of π.

Using our log law, we can manipulate five multiplied by the natural logarithm of π₯ to be equal to the natural logarithm of π₯ raised to the fifth power. This means that π raised to the power of five multiplied by the natural logarithm of π₯ is equal to π raised to the power of the natural logarithm of π₯ to the fifth power. And we know that π raised to the power of the natural logarithm of π is just equal to π. And that means that π raised to the power of the natural logarithm of π₯ to the fifth power is just equal to π₯ to the fifth power.

Weβre now ready to evaluate the integral given to us in the question. First, we know that π raised to the power of five multiplied by the natural logarithm of π₯ is equal to π₯ to the fifth power. So the first term in our integrand will be π₯ to the fifth power and then we subtract π raised to the power of seven π₯. We can now integrate each term separately. We know that if π is not equal to negative one, then the integral of π₯ to the πth power with respect to π₯ is equal to π₯ to the power of π plus one all divided by π plus one plus the constant of integration π. Using this, we have the integral of π₯ to the fifth power with respect to π₯ is equal to π₯ to the power of six divided by six plus the constant of integration π one.

Next, we know that for a constant π and a constant π not equal to zero, the integral of π multiplied by π to the power of π multiplied by π₯ with respect to π₯ is equal to π divided by π multiplied by π to the power of π multiplied by π₯ plus the constant of integration π. Using this gives us that the integral of negative π to the power of seven π₯ with respect to π₯ is equal to negative one-seventh multiplied by π to the power of seven π₯ plus the constant of integration we will call π two. Now since both π one and π two are constants, we could combine both of these into a new constant, which we will call π.

Hence, we have shown that the integral of π to the power of five multiplied by the natural logarithm of π₯ minus π to the power of seven multiplied by π₯ with respect to π₯ is equal to one-sixth multiplied by π₯ to the sixth power minus one-seventh multiplied by π to the power of seven π₯ plus the constant of integration π.