Video: Finding the Integration of a Rational Function by Using Integration by Substitution

Determine ∫ (π‘₯⁡ + 1)/(π‘₯⁢ + 6π‘₯)⁸ dπ‘₯.

02:56

Video Transcript

Determine the indefinite integral of π‘₯ to the fifth power plus one over π‘₯ to the sixth power plus six π‘₯ to the eighth power with respect to π‘₯.

Now this is not an expression that’s nice to integrate using our standard rules for finding the antiderivative. Instead, we can use the substitution rule for integration. Now this says that if 𝑒 is some function of π‘₯, 𝑒 equals 𝑔 of π‘₯ is a differentiable function whose range is an interval 𝑖 and 𝑓 is continuous on that interval. Then the integral of 𝑓 of 𝑔 of π‘₯ times 𝑔 prime of π‘₯ with respect to π‘₯ is equal to the integral of 𝑓 of 𝑒 with respect to 𝑒.

We try to choose 𝑒 to be some factor of the integrand whose derivative also occurs or some scalar multiple of it. And if that’s not possible, we try choosing 𝑒 to be some more complicated part of the integrand, like the inner function in a composite function. Now if we have a look at the inner part in our composite function on the denominator of our fraction, that’s π‘₯ to the sixth power plus six π‘₯. We see that the numerator of our fraction is a scalar multiple of the derivative of this expression. So we’re going to let 𝑒 be equal to π‘₯ to the sixth power plus six π‘₯. Then we can say that d𝑒 by dπ‘₯ is equal to six π‘₯ to the fifth power plus six.

Now what the substitution rule also says is that it’s permissible to operate with dπ‘₯ and d𝑒 after integral signs as if they were differentials. In other words, we’re allowed to treat d𝑒 by dπ‘₯ a little bit like a fraction in this scenario. So we can rewrite this as d𝑒 equals six π‘₯ to the fifth power plus six dπ‘₯.

Now I’m going to divide through by six. And we find that a sixth d𝑒 equals π‘₯ to the fifth power plus one dπ‘₯. And this is great because we can now replace π‘₯ to the sixth power plus six π‘₯ with 𝑒. And we can replace π‘₯ to the fifth power plus one dπ‘₯ with a sixth d𝑒. And so our integral becomes one over six 𝑒 to the eighth power d𝑒.

We’ll take the constant factor of one-sixth outside of our integral. And then we’ll rewrite one over 𝑒 to the eighth power as 𝑒 to the power of negative eight. Then when we integrate a polynomial term of the form π‘Žπ‘₯ to the 𝑛th power for values of 𝑛 not equal to negative one, we add one to the exponent and then divide by that value. And so the integral of 𝑒 to the power of negative eight is 𝑒 to the power of negative seven over negative seven plus the constant of integration π‘Ž.

We’re now going to replace 𝑒 with π‘₯ to the sixth power plus six π‘₯. So we find our integral is equal to one-sixth times negative one-seventh times π‘₯ to the sixth power plus six π‘₯ to the power of negative seven plus a constant of integration π‘Ž. We’re now going to distribute our parentheses. Now doing so changes our constant of integration. So let’s call that capital 𝐢. We can also write π‘₯ to the sixth power plus six π‘₯ to the power of negative seven as one over π‘₯ to the sixth power plus six π‘₯ to the power of seven. And so we find that our integral is equal to negative one over 42 times π‘₯ to the sixth power plus six π‘₯ to the seventh power plus 𝐢.

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