Video: Finding the Integration of a Function That Is the Sum of a Reciprocal and Quadratic Functions

Determine ∫ ((5𝑒⁴/π‘₯) + 5π‘₯Β² ln 6) dπ‘₯.

02:38

Video Transcript

Determine the integral of five 𝑒 to the fourth power over π‘₯ plus five π‘₯ squared times the natural logarithm of six with respect to π‘₯.

In this question, we’re asked to evaluate an integral. And we can see that our integral is the sum of two functions. This means we can try and do this term by term. And at first we might be worried about how to do this. For example, in our first term, we can see we have a factor of 𝑒 to the fourth power. And in our second term we have the natural logarithm of six. But both of these are constants; they’re not varying as the value of π‘₯ varies. In fact, only two parts of this expression vary as π‘₯ varies. We can see we divide by π‘₯ in our first term and we have π‘₯ squared in our second term.

So in our first term, five 𝑒 to the fourth power is a constant and, in our second term, five times the natural logarithm of six is also a constant. So in fact, we know how to integrate both of these expressions separately. To integrate the first term, we need to recall the following integral result for reciprocal functions. For any real constant π‘Ž, the integral of π‘Ž over π‘₯ with respect to π‘₯ is equal to π‘Ž times the natural logarithm of the absolute value of π‘₯ plus the constant of integration 𝐢. And in our first term, we can see the value of the constant π‘Ž is five 𝑒 to the fourth power.

So by setting the value of π‘Ž equal to five 𝑒 to the fourth power in our integral rule, we can integrate our first term to give us five 𝑒 to the fourth power times the natural logarithm of the absolute value of π‘₯. And it’s worth pointing out we only need to add one constant of integration at the end of this expression. So we won’t add this yet. Now, to integrate our second term, we’ll use the power rule for integration. We recall this tells us, for any real values π‘Ž and 𝑛, where 𝑛 is not equal to negative one, the integral of π‘Žπ‘₯ to the 𝑛th power with respect to π‘₯ is equal to π‘Ž times π‘₯ to power of 𝑛 plus one divided by 𝑛 plus one plus the constant of integration 𝐢.

And we add one to our exponent of π‘₯ and then divide by this new exponent. This time, our value, the constant π‘Ž, is equal to five times the natural logarithm of six, and our exponent 𝑛 is equal to two. So we can use this to integrate our second term. We set π‘Ž equal to five times the natural logarithm of six and 𝑛 equal to two. This gives us five times the natural logarithm of six multiplied by π‘₯ cubed all over three. And of course, we need to add our constant of integration 𝐢.

And in fact, we could leave our answer like this. However, we’ll rearrange this slightly. In our second term, we’ll take the constant factor of the natural logarithm of six and write this as the end of our second term. And doing this gives us our final answer.

Therefore, we were able to show the integral of five 𝑒 to the fourth power over π‘₯ plus five π‘₯ squared times the natural logarithm of six with respect to π‘₯ is equal to five 𝑒 to the fourth power times the natural logarithm of the absolute value of π‘₯ plus five π‘₯ cubed over three times the natural logarithm of six plus our constant of integration 𝐢.

Nagwa uses cookies to ensure you get the best experience on our website. Learn more about our Privacy Policy.