### Video Transcript

Evaluate the integral from eight to seven plus π of five divided by π₯ minus seven times the natural logarithm of π₯ minus seven all squared plus one all cubed with respect to π₯.

In this question, weβre asked to evaluate a very complicated-looking integral. And the first thing we should ask is, can we simplify our integrand? And in this case, we can. For example, by using the power rule for logarithms, we can simplify the natural logarithm of π₯ minus seven all squared. We can simplify this to give us two times the natural logarithm of π₯ minus seven. And at first, it might be hard to see how this helps us evaluate our integral. We need to notice something about our integrand.

At the front of this expression, we have five divided by π₯ minus seven. And this is a scalar multiple of the derivative of the natural logarithm of π₯ minus seven with respect to π₯. And because a scalar multiple of the derivative of this expression appears in our integrand, we could try using integration by substitution. So we want to try the substitution π’ is equal to the natural logarithm of π₯ minus seven. We differentiate both sides of our substitution with respect to π₯ to give us dπ’ by dπ₯ is equal to one over π₯ minus seven. Then, although dπ’ by dπ₯ is not a fraction, when weβre using integration by substitution, we can treat it a little bit like a fraction.

This gives us the equivalent statement in terms of differentials, dπ’ is equal to one over π₯ minus seven dπ₯. And because the expression five divided by π₯ minus seven is what appears in our integrand, weβll multiply both sides of this equation through by five. So we have five dπ’ is equal to five over π₯ minus seven dπ₯. Weβre almost ready to try and evaluate this by substitution. However, remember, this is a definite integral. So we need to find the new limits of integration. To do this, we substitute these values of π₯ into our expression for π’.

Letβs start with the new upper limit of integration. We substitute π₯ is equal to seven plus π into our expression for π’. This gives us π’ is equal to the natural logarithm of seven plus π minus seven. This, of course, simplifies to give us the natural logarithm of π, which we know is equal to one. We can do the same to find the new lower limit of integration. We substitute π₯ is equal to eight into our expression for π’. This gives us the natural logarithm of eight minus seven, which simplifies to give us the natural logarithm of one, which is of course equal to zero. Weβre now ready to try and evaluate our integral by using our substitution.

Remember, we simplified our integrand by using the power rule for logarithms. So we simplified our integral to give us the following expression, and weβre now ready to apply our substitution. First, by using the substitution π’ is equal to the natural logarithm of π₯ minus seven, we showed the new lower limit of integration was zero and the new upper limit of integration was one. Next, we showed, in terms of differentials, five divided by π₯ minus seven dπ₯ is equivalent to five dπ’, so we can just replace this with five dπ’. Finally, remember the natural logarithm of π₯ minus seven is equal to π’. So we can replace this by π’, giving us the following expression. And we can simplify this slightly. We can take the constant factor of five outside of our integral.

So now we need to evaluate five times the integral from zero to one of two π’ plus one all cubed with respect to π’. And thereβs a few different ways of doing this. For example, we could use substitution once again. However, because weβre only cubing a binomial, we can just distribute over our parentheses. Weβll distribute the cube by using the binomial formula. Our first term will be three choose zero multiplied by two π’ all cubed. And we can evaluate this. Three choose zero is equal to one, and two π’ all cubed is equal to eight π’ cubed. So our first term is eight π’ cubed. By using the binomial formula, our second term will be three choose one times two π’ all squared multiplied by one.

And once again, we can evaluate this. First, three choose one is equal to three. Next, multiplying by one doesnβt change the value. Finally, two π’ all squared is equal to four π’ squared. So our second term simplifies to give us three times four π’ squared which is, of course, equal to 12π’ squared. We can use the binomial formula to find the third term in our expansion. Itβs equal to three choose two times two π’ multiplied by one squared. And of course, we can simplify this. First, three choose two is equal to three. Next, one squared doesnβt change the value of our expression. Finally, three times two π’ is equal to six π’.

So our third term is six π’, and our final term will be three choose three multiplied by one cubed. And both of these are equal to one, so this simplifies to just give us one. So by distributing the cube over our parentheses, we were able to rewrite our integral in the following form. And we see we can evaluate this integral by using the power rule for integration. On each of these terms, we need to add one to our exponent of π’ and then divide by this new exponent. This gives us the following expression. And of course, we can simplify this. First, eight divided by four is equal to two. Next, 12 divided by three is equal to four. Then, six divided by two is equal to three.

So now we have five times two π’ to the fourth power plus four π’ cubed plus three π’ squared plus π’ evaluated at the limits of integration π’ is equal to zero and π’ is equal to one. Now, all thatβs left to do is evaluate this at the limits of integration. However, before we do this, we can notice something. If we were to substitute π’ is equal to zero into our antiderivative, we can see every single term evaluates to give us zero. So we only need to evaluate our antiderivative at one, giving us five times two multiplied by one to the fourth power plus four times one cubed plus three times one squared plus one. Then, all we have to do is calculate this expression and we see itβs equal to 50.

Therefore, by using integration by substitution, we were able to show the integral from eight to seven plus π of five divided by π₯ minus seven times the natural logarithm of π₯ minus seven all squared plus one all cubed with respect to π₯ is equal to 50.