Video Transcript
Calculate the integral of one over
𝑡 plus six 𝐢 plus one over two 𝑡 plus five all raised to the fourth power 𝐣 plus
five 𝑡 to the fourth power divided by 𝑡 to the fifth power plus five 𝐤 with
respect to 𝑡.
We’re asked to evaluate the
integral of a vector-valued function. And to do this, we need to recall
that we do this component-wise. In other words, we need to evaluate
the integral of each component function separately. Let’s start by evaluating the
integral of our first component function. That’s the integral of one divided
by 𝑡 plus six with respect to 𝑡. There’s several different ways of
evaluating this integral. For example, we could substitute 𝑢
is equal to 𝑡 plus six. And this would work. However, the easiest way is to
recall the following integral result.
We know the integral of 𝑓 prime of
𝑡 divided by 𝑓 of 𝑡 with respect to 𝑡 is equal to the natural logarithm of the
absolute value of 𝑓 of 𝑡 plus a constant of integration 𝑐. And we can see this is exactly what
we have in our integral. The function 𝑓 of 𝑡 in our
denominator would be 𝑡 plus six. And then we can see the derivative
of 𝑡 plus six with respect to 𝑡 would be the coefficient of 𝑡, which is one,
which is exactly equal to our numerator. So we can evaluate this integral by
setting our function 𝑓 of 𝑡 to be 𝑡 plus six. This gives us the natural logarithm
of the absolute value of 𝑡 plus six plus a constant of integration we’ll call
lowercase 𝑎.
We now want to evaluate the
integral of our second component function. That means we want to evaluate the
integral of one divided by two 𝑡 plus five all raised to the fourth power with
respect to 𝑡. This time, we’ll do this by using a
substitution. We’re going to use the substitution
𝑢 is equal to two 𝑡 plus five. We then need to differentiate both
sides of this expression with respect to 𝑡. We get d𝑢 by d𝑡 will be the
derivative of two 𝑡 plus five with respect to 𝑡. This is a linear function. So its derivative with respect to
𝑡 will be the coefficient of 𝑡, which is two.
Now although we know d𝑢 by d𝑡 is
not a fraction, we can treat it a little bit like a fraction when we’re doing
integration by substitution. This gives us the equivalent
statement in terms of differentials: d𝑢 is equal to two d𝑡. And one step that can be helpful is
to notice in our integral we don’t have the expression two d𝑡. Instead, we have one d𝑡. So we’ll divide both sides of our
equation in terms of differentials by two. This gives us one-half d𝑢 is equal
to one d𝑡.
We’re now ready to evaluate our
integral by substitution. First, we use our substitution 𝑢
is equal to two 𝑡 plus five to rewrite our denominator as 𝑢 to the fourth
power. Then, all we need to do is use our
equation in terms of differentials to rewrite our integral as the integral of
one-half over 𝑢 to the fourth power with respect to 𝑢. And we can simplify this
integral. First, we’ll take the constant
factor of one-half outside of our integral.
Next, by using our laws of
exponents, we’ll rewrite one over 𝑢 to the fourth power as 𝑢 to the power of
negative four. This means we need to evaluate the
integral of one-half times 𝑢 to the power of negative four with respect to 𝑢. We can do this by using the power
rule for integration. We want to add one to our exponent
of 𝑢, giving us a new exponent of negative three, and then divide by this new
exponent. This gives us 𝑢 to the power of
negative three divided by two times negative three. And of course, we need to add a
constant of integration. We’ll call our constant of
integration lowercase 𝑏.
And we can simplify this. In our denominator, two multiplied
by negative three is equal to negative six. And instead of multiplying by 𝑢 to
the power of negative three, we can instead divide by 𝑢 cubed. This gives us negative one over six
𝑢 cubed plus 𝑏. But remember, we want to give our
expression in terms of 𝑡. To do this, we’re going to use the
substitution 𝑢 is equal to two 𝑡 plus five. Doing this, we get the integral of
our second component function with respect to 𝑡 is equal to negative one divided by
six times two 𝑡 plus five all cubed plus 𝑏.
We now want to find the integral of
our third component function. We’ll start by clearing some
space. We now want to find the integral of
our third component function. That’s the integral of five 𝑡 to
the fourth power divided by 𝑡 to the fifth power plus five with respect to 𝑡. Once again, we can do this by
substitution. We could use the substitution 𝑢 is
equal to 𝑡 to the fifth power plus five. And the reason we might choose this
substitution is the derivative of this expression appears in our numerator.
And this would work. However, we’ve already done this in
the general case, and we’ve written our formula earlier. In this case, our function 𝑓 of 𝑡
would be 𝑡 to the fifth power plus five. And then by using the power rule
for differentiation, we know that 𝑓 prime of 𝑡 would be five 𝑡 to the fourth
power, which is our numerator. So we can just integrate this
expression with respect to 𝑡 by setting 𝑓 of 𝑡 to be 𝑡 to the fifth power plus
five. This gives us the natural logarithm
of the absolute value of 𝑡 to the fifth power plus five plus a constant of
integration we’ll call lowercase 𝑐.
Now that we’ve integrated all three
of the component functions with respect to 𝑡, we’re ready to evaluate the integral
of the vector-valued function given to us in the question. First, we showed the integral of
the first component function is the natural logarithm of the absolute value of 𝑡
plus six plus 𝑎. Next, we showed the integral of our
second component function is negative one divided by six times two 𝑡 plus five all
cubed plus 𝑏. And it’s worth pointing out here we
took the negative one outside of our vector and used the fact that lowercase 𝑏 is a
constant of integration. Finally, we were able to show the
integral of our third component function was the natural logarithm of the absolute
value of 𝑡 to the fifth power plus five plus our constant of integration lowercase
𝑐.
So when we integrate our
vector-valued function, we just write the integrals as the new component
functions. This gives us the following
expression for our vector-valued function. And we could leave our answer like
this. However, there’s something worth
noticing. If we were to distribute our
parentheses, we would notice we would get the term 𝑎𝐢 plus 𝑏𝐣 plus 𝑐𝐤. We know that lowercase 𝑎, 𝑏, and
𝑐 are all constants of integration. So 𝑎𝐢, 𝑏𝐣, and 𝑐𝐤 would just
be constant vectors. So we could combine all of these
into a new vector. We’ll combine these into a new
constant vector we’ll call capital 𝐂. And this gives us our final
answer.
Therefore, we were able to show the
integral of one over 𝑡 plus six 𝐢 plus one over two 𝑡 plus five all raised to the
fourth power 𝐣 plus five 𝑡 to the fourth power over 𝑡 to the fifth power plus
five 𝐤 with respect to 𝑡 is equal to the natural logarithm of the absolute value
of 𝑡 plus six 𝐢 minus one over six times two 𝑡 plus five all cubed 𝐣 plus the
natural logarithm of the absolute value of 𝑡 to the fifth power plus five 𝐤 plus a
constant vector 𝐂.
