Video: Integrate Exponential Functions

Find ∫(𝑒^(5 ln π‘₯) βˆ’ 𝑒^(7π‘₯)) dπ‘₯.

02:54

Video Transcript

Find the integral of 𝑒 raised to the power of five multiplied by the natural logarithm of π‘₯ minus 𝑒 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 𝑒 raised to the power of five multiplied by the natural logarithm of π‘₯ is a composition of functions we don’t know how to integrate. So we’ll need to manipulate this. To do this, we’re going to use the following two laws. The first tells us that 𝑒 raised to the power of the natural logarithm of π‘Ž is just equal to π‘Ž. And the second one tells us that 𝑏 multiplied by the natural logarithm of π‘Ž is equal to the natural logarithm of π‘Ž raised to the power of 𝑏.

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 𝑒 raised to the power of five multiplied by the natural logarithm of π‘₯ is equal to 𝑒 raised to the power of the natural logarithm of π‘₯ to the fifth power. And we know that 𝑒 raised to the power of the natural logarithm of π‘Ž is just equal to π‘Ž. And that means that 𝑒 raised to the power of the natural logarithm of π‘₯ to the fifth power is just equal to π‘₯ to the fifth power.

We’re now ready to evaluate the integral given to us in the question. First, we know that 𝑒 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 𝑒 raised to the power of seven π‘₯. We can now integrate each term separately. We know that if 𝑛 is not equal to negative one, then the integral of π‘₯ to the 𝑛th power with respect to π‘₯ is equal to π‘₯ to the power of 𝑛 plus one all divided by 𝑛 plus one plus the constant of integration 𝑐. 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 𝑐 one.

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

Hence, we have shown that the integral of 𝑒 to the power of five multiplied by the natural logarithm of π‘₯ minus 𝑒 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 𝑒 to the power of seven π‘₯ plus the constant of integration 𝑐.

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