Question Video: Finding the Integration of a Rational Function Using Integration by Substitution Mathematics • Third Year of Secondary School

Video Transcript

Determine the integral of two 𝑥 divided by seven 𝑥 squared plus one with respect to 𝑥.

We’re asked to evaluate the integral of the quotient of two polynomials. That’s a rational function. And there’s a few different ways we know how to try and integrate rational functions. For example, we might want to try using partial fractions. However, in this case, our denominator doesn’t factor, so this won’t work. We could also try something like substitution. However, in actual fact, there’s a simpler method.

Let’s call the function in our denominator 𝑓 of 𝑥. So that’s seven 𝑥 squared plus one. We want to check what 𝑓 prime of 𝑥 is. We can do this by using the power rule for differentiation. We get that 𝑓 prime of 𝑥 is 14𝑥. And we can see that 14𝑥 is a linear multiple of our numerator. So we can integrate this by using one of our integral rules. We need to recall the following. The integral of 𝑓 prime of 𝑥 over 𝑓 of 𝑥 with respect to 𝑥 is equal to the natural logarithm of the absolute value of 𝑓 of 𝑥 plus the constant of integration 𝐶.

Now in this case, our expression is not quite in this form. If we set 𝑓 of 𝑥 to be our denominator, we’ve shown the derivative of seven 𝑥 squared plus one with respect to 𝑥 is 14𝑥. However, our numerator is two 𝑥. We need our numerator to be 14𝑥, so what we could do is multiply this by seven and then divide our entire integral by seven. Multiplying by seven and dividing by seven is the same as multiplying by one, so it won’t change our value. However, now we can see our numerator will be 14𝑥. That’s 𝑓 prime of 𝑥.

So now we can evaluate this integral by using our integral rule. We’ll set 𝑓 of 𝑥 to be seven 𝑥 squared plus one. This gives us one-seventh times the natural logarithm of the absolute value of seven 𝑥 squared plus one plus a constant of integration we’ll call 𝐴. And we can simplify this expression. First, we’ll distribute one-seventh over our parentheses. This gives us one-seventh times the natural logarithm of the absolute value of seven 𝑥 squared plus one plus 𝐴 over seven. And remember, 𝐴 is just a constant, so 𝐴 over seven is also a constant. We can just call this a new constant called 𝐶.

And by doing this, we get our final answer. Therefore, we were able to show the integral of two 𝑥 over seven 𝑥 squared plus one with respect to 𝑥 is equal to one-seventh times the natural logarithm of the absolute value of seven 𝑥 squared plus one plus our constant of integration 𝐶.