### Video Transcript

Determine the integral with respect

to 𝑥 of 25𝑥 squared plus 40𝑥 plus 17 all divided by the square root of five 𝑥

plus four.

This looks like quite a complex

rational function to integrate. We have a quadratic in the

numerator and a root function in the denominator. But we can make things a little

more manageable using the technique, integration by substitution. This means replacing a function of

𝑥 with 𝑢 and so finding d𝑥 in terms of d𝑢. We then integrate with respect to

𝑢 and finally substitute back in our function of 𝑥 for 𝑢. So what’s a good substitution to

make here. Well, the square root in our

denominator contains a linear function of 𝑥. Differentiating such a function

gives us a constant, and this makes things simpler when we find d𝑥 in terms of

d𝑢.

So let’s try this with the

substitution 𝑢 is equal to five 𝑥 plus four. Differentiating 𝑢 with respect to

𝑥, we find d𝑢 by d𝑥 is equal to five. And although we know that d𝑢 by

d𝑥 is not a fraction as such, in this context, we can treat it a little like a

fraction so that dividing both sides by five and multiplying by d𝑥, we find one

over five d𝑢 is equal to d𝑥. So we have 𝑢 is equal to five 𝑥

plus four and in terms of our integral by taking square roots, we have the square

root of 𝑢 is equal to the square root of five 𝑥 plus four. And this is our denominator.

So now, let’s consider how we might

apply this substitution to our numerator. First solving 𝑢 for 𝑥, that is,

by subtracting four from both sides and dividing by five, we find 𝑢 minus four over

five is equal to 𝑥. Squaring both sides, we have 𝑢

minus four over five all squared is equal to 𝑥 squared. And squaring our numerator and

denominator on the left-hand side gives us 𝑢 minus four squared over 25 is equal to

𝑥 squared. And now multiplying through by 25,

on our left-hand side we have 𝑢 minus four squared and on our right-hand side 25𝑥

squared. So 𝑢 minus four squared is 25𝑥

squared. And this is the first term in our

numerator

And now looking again at our

substitution, 𝑢 is equal to five 𝑥 plus four, we’d like to find the second term in

the numerator, that’s 40𝑥, in terms of 𝑢. To find this, we again subtract

four from both sides, giving 𝑢 minus four is equal to five 𝑥. And now, we multiply both sides by

eight. Distributing our parentheses on the

left-hand side then, we have eight 𝑢 minus 32 is equal to 40𝑥. And this is the second term in our

numerator.

So now we can write our numerator

in terms of 𝑢. And that is 𝑢 minus four squared,

which is 25𝑥 squared plus eight 𝑢 minus 32 which is 40𝑥. And we add our constant 17. Distributing our parentheses on the

right-hand side, we now have 𝑢 squared minus eight 𝑢 plus 16 plus eight 𝑢 minus

32 plus 17. And collecting like terms, we’re

left with 𝑢 squared plus one. And so in terms of 𝑢, our

numerator is simply 𝑢 squared plus one.

So now we have d𝑥 in both our

numerator and denominator in terms of 𝑢. We can take the one over five,

which is a constant, multiplying d𝑢 in front of our integral sign. And so with the substitution 𝑢 is

equal to five 𝑥 plus four, our integral with respect to 𝑥 has become one over five

times the integral of 𝑢 squared plus one over the square root of 𝑢 with respect to

𝑢.

We can use the laws of exponents to

split our integrand since this is 𝑢 squared over the square root of 𝑢 plus one

over the square root of 𝑢. And recalling that the square root

of 𝑢 is actually 𝑢 to the power of a half, we can use our law for negative

exponents that tells us one over 𝑎 raised to the power 𝑝 is 𝑎 raise to the power

negative 𝑝, which gives us 𝑢 squared multiplied by 𝑢 raised to the power negative

a half plus 𝑢 raised to the power negative a half. We can then use the product law for

exponents, which tells us that 𝑎 raised to the power 𝑝 multiplied by 𝑎 raised to

the power 𝑞 is 𝑎 raised to the power 𝑝 plus 𝑞 so that our first term is then 𝑢

raised to the power three over two. And our integrand is then 𝑢 raised

the power three over two plus 𝑢 raised the power negative a half.

And now, applying the property of

linearity to our integral, we have one over five times the integral of 𝑢 raised to

the power three over two with respect to 𝑢 plus one over five times the integral of

𝑢 raised to the power negative a half with respect to 𝑢. So now, we can apply the power rule

for integrals to both terms. And this tells us that for 𝑛 not

equal to negative one, the integral 𝑣 raised to the power 𝑛 with respect to 𝑣 is

equal to 𝑣 raised to the power 𝑛 plus one all divided by 𝑛 plus one plus a

constant of integration 𝐶. Applying this to both of our

integrals, we add one to each of our exponents and divide by the sum of our exponent

and one. And we add a constant of

integration for each of our integrals.

And now, since three over two plus

one is equal to five over two and negative one over two plus one is equal to one

over two, we now have one over five multiplied by 𝑢 raised the power five over two

divided by five over two plus one over five multiplied by 𝑢 raised to the power of

a half divided by a half. And we collect our two constants of

integration 𝐶 one and 𝐶 two.

And now recalling that to divide by

a fraction, we flip it and multiply, we have two over 25 multiplied by 𝑢 raised to

the power five over two plus two over five multiplied by 𝑢 raised to the power of a

half plus our constant of integration 𝐶 one plus 𝐶 two, which we’ve called 𝐶. And we can do this since these are

arbitrary constants. And so, we found our integral in

terms of 𝑢 and so finally recalling our substitution 𝑢 is equal to five 𝑥 plus

four. Substituting 𝑢 into our result,

the integral of 25𝑥 squared plus 40𝑥 plus 17 all over the square root of five 𝑥

plus four with respect to 𝑥 is equal to two over 25 multiplied by five 𝑥 plus four

raised to the power five over two plus two over five multiplied by the square root

of five 𝑥 plus four plus the constant of integration 𝐶.