### Video Transcript

Determine the integral with respect
to π₯ of 25π₯ squared plus 40π₯ plus 17 all divided by the square root of five π₯
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
π₯ with π’ and so finding dπ₯ in terms of dπ’. We then integrate with respect to
π’ and finally substitute back in our function of π₯ for π’. So whatβs a good substitution to
make here. Well, the square root in our
denominator contains a linear function of π₯. Differentiating such a function
gives us a constant, and this makes things simpler when we find dπ₯ in terms of
dπ’.

So letβs try this with the
substitution π’ is equal to five π₯ plus four. Differentiating π’ with respect to
π₯, we find dπ’ by dπ₯ is equal to five. And although we know that dπ’ by
dπ₯ 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 dπ₯, we find one
over five dπ’ is equal to dπ₯. So we have π’ is equal to five π₯
plus four and in terms of our integral by taking square roots, we have the square
root of π’ is equal to the square root of five π₯ plus four. And this is our denominator.

So now, letβs consider how we might
apply this substitution to our numerator. First solving π’ for π₯, that is,
by subtracting four from both sides and dividing by five, we find π’ minus four over
five is equal to π₯. Squaring both sides, we have π’
minus four over five all squared is equal to π₯ squared. And squaring our numerator and
denominator on the left-hand side gives us π’ minus four squared over 25 is equal to
π₯ squared. And now multiplying through by 25,
on our left-hand side we have π’ minus four squared and on our right-hand side 25π₯
squared. So π’ minus four squared is 25π₯
squared. And this is the first term in our
numerator

And now looking again at our
substitution, π’ is equal to five π₯ plus four, weβd like to find the second term in
the numerator, thatβs 40π₯, in terms of π’. To find this, we again subtract
four from both sides, giving π’ minus four is equal to five π₯. And now, we multiply both sides by
eight. Distributing our parentheses on the
left-hand side then, we have eight π’ minus 32 is equal to 40π₯. And this is the second term in our
numerator.

So now we can write our numerator
in terms of π’. And that is π’ minus four squared,
which is 25π₯ squared plus eight π’ minus 32 which is 40π₯. And we add our constant 17. Distributing our parentheses on the
right-hand side, we now have π’ squared minus eight π’ plus 16 plus eight π’ minus
32 plus 17. And collecting like terms, weβre
left with π’ squared plus one. And so in terms of π’, our
numerator is simply π’ squared plus one.

So now we have dπ₯ in both our
numerator and denominator in terms of π’. We can take the one over five,
which is a constant, multiplying dπ’ in front of our integral sign. And so with the substitution π’ is
equal to five π₯ plus four, our integral with respect to π₯ has become one over five
times the integral of π’ squared plus one over the square root of π’ with respect to
π’.

We can use the laws of exponents to
split our integrand since this is π’ squared over the square root of π’ plus one
over the square root of π’. And recalling that the square root
of π’ is actually π’ to the power of a half, we can use our law for negative
exponents that tells us one over π raised to the power π is π raise to the power
negative π, which gives us π’ squared multiplied by π’ raised to the power negative
a half plus π’ raised to the power negative a half. We can then use the product law for
exponents, which tells us that π raised to the power π multiplied by π raised to
the power π is π raised to the power π plus π so that our first term is then π’
raised to the power three over two. And our integrand is then π’ raised
the power three over two plus π’ 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 π’ raised to
the power three over two with respect to π’ plus one over five times the integral of
π’ raised to the power negative a half with respect to π’. So now, we can apply the power rule
for integrals to both terms. And this tells us that for π not
equal to negative one, the integral π£ raised to the power π with respect to π£ is
equal to π£ raised to the power π plus one all divided by π plus one plus a
constant of integration πΆ. 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 π’ raised the power five over two
divided by five over two plus one over five multiplied by π’ raised to the power of
a half divided by a half. And we collect our two constants of
integration πΆ one and πΆ two.

And now recalling that to divide by
a fraction, we flip it and multiply, we have two over 25 multiplied by π’ raised to
the power five over two plus two over five multiplied by π’ raised to the power of a
half plus our constant of integration πΆ one plus πΆ two, which weβve called πΆ. And we can do this since these are
arbitrary constants. And so, we found our integral in
terms of π’ and so finally recalling our substitution π’ is equal to five π₯ plus
four. Substituting π’ into our result,
the integral of 25π₯ squared plus 40π₯ plus 17 all over the square root of five π₯
plus four with respect to π₯ is equal to two over 25 multiplied by five π₯ plus four
raised to the power five over two plus two over five multiplied by the square root
of five π₯ plus four plus the constant of integration πΆ.