### Video Transcript

Evaluate the integral from one to
two of eight π‘ divided by 10π‘ squared plus five with respect to π‘.

Weβre asked to calculate the
integral of a rational function, and weβve seen many methods for doing this. We need to decide which one weβre
going to apply in this case. We can notice the derivative of our
denominator 10π‘ squared plus five with respect to π‘ is equal to 20π‘. And we see this is a scalar
multiple of our numerator. Then if we set our denominator
equal to the function π of π‘, we can notice weβre asked to integrate a scalar
multiple π of the derivative of π divided by π of π‘ with respect to π‘. In fact, this is a special case of
an integration by substitution rule weβve seen before when we set our function π of
π‘ equal to one divided by π‘.

We recall if our function π prime
is continuous on the closed interval from π to π and our function π is continuous
on the range of π’ is equal to π of π‘. Then by using integration by
substitution with the substitution π’ is equal to π of π‘. We have the integral from π to π
of π prime of π‘ multiplied by π evaluated at π of π‘ with respect to π‘ is equal
to the integral from π evaluated at π to π evaluated at π of π evaluated at π’
with respect to π’. And weβve already shown we can
write our integral in this manner. In fact, since π is a constant, we
can take it outside of our integral.

So we want to use the substitution
π’ is equal to π of π‘ which is 10π‘ squared plus five. This gives us that our function π
prime of π‘ is equal to 20π‘ which is continuous on all of the real numbers, which
means, in particular, it must be continuous on the closed interval of our integral,
which in this case is the closed interval from one to two. Next, we need to check that our
function π, which in this case is one divided by π‘, is continuous on the range of
π’ is equal to π of π‘. We find the range of π of π‘ by
evaluating the function at the limits of our integral. Thatβs π of two and π of one.

We have that π evaluated at two is
equal to 10 multiplied by two squared plus five, which we can evaluate to give us
45. Similarly, we can evaluate π at
one to give us 10 multiplied by one squared plus five, which is equal to 15. So we need our function π to be
continuous on the closed interval from 15 to 45. And since our function π of π‘ is
just equal to one divided by π‘ is continuous on any interval which does not include
zero, so, in particular, itβs continuous on the range of π of π‘. So weβre now ready to use
integration by substitution by using π’ is equal to 10π‘ squared plus five.

Differentiating both sides of this
equation gives us that dπ’ dπ‘ is equal to 20π‘. And although dπ’ by dπ‘ is not a
fraction, when using integration by substitution, it does behave a little bit like a
fraction. This gives us the equivalent
statement dπ’ is equal to 20π‘ dπ‘. Weβre almost ready to evaluate this
integral. We just need to rewrite our
numerator as a scalar multiple of the derivative of π’. And we see if our constant π is
equal to eight divided by 20, then eight divided by 20 multiplied by 20π‘ is equal
to eight π‘. And we can take this constant of
eight divided by 20 outside of our integral. We now have that our denominator of
10π‘ squared plus five is equal to π’ and 20π‘ dπ‘ is equal to dπ’.

Finally, the new limits of our
integral will be from π evaluated at one to π evaluated at two. Thatβs from 15 to 45. So the substitution π’ is equal to
10π‘ squared plus five gave us that our integral is equal to eight divided by 20
multiplied by the integral from 15 to 45 of one divided by π’ with respect to
π’. Now, we can evaluate our integral
by using the fact that the integral of one divided by π₯ with respect to π₯ is equal
to the natural logarithm of the absolute value of π₯ plus a constant of integration
πΆ. We simplify eight divided by 20 to
give us two divided by five. Then we integrate one divided by π’
to give us the natural logarithm of the absolute value of π’, while we ignore our
constant of integration because it will cancel when we evaluate at the limits of our
integral π’ is equal to 15 and π’ is equal to 45.

Evaluating at the limits of our
integral gives us two-fifths multiplied by the natural logarithm of the absolute
value of 45 minus the natural logarithm of the absolute value of 15. Both 45 and 15 are positive. So their absolute values are just
equal to themselves. So we can remove the absolute
value. Finally, we can simplify this by
using the quotient rule for logarithms, which tells us the log of π minus the log
of π is equal to the log of π divided by π. The difference between two
logarithms is equal to the quotient of the logarithm.

So the natural logarithm of 45
minus the natural logarithm of 15 can be simplified to the natural logarithm of 45
divided by 15. And 45 over 15 simplifies to three,
giving us that our integral is two times the natural logarithm of three divided by
five. Therefore, by using integration by
substitution, weβve shown that the integral from one to two of eight π‘ divided by
10π‘ squared plus five with respect to π‘ is equal to two times the natural
logarithm of three all divided by five.