Question Video: Finding the Integration of a Function Involving a Root Function Using Integration by Substitution | Nagwa Question Video: Finding the Integration of a Function Involving a Root Function Using Integration by Substitution | Nagwa

# Question Video: Finding the Integration of a Function Involving a Root Function Using Integration by Substitution Mathematics • Third Year of Secondary School

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Determine β«(25π₯Β² + 40π₯ + 17)/β(5π₯ + 4) dπ₯.

06:11

### 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 whatβs 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 letβs 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, letβs 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, weβd like to find the second term in the numerator, thatβs 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, weβre 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 πΆ. 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 πΆ one and πΆ 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 πΆ one plus πΆ two, which weβve called πΆ. 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 πΆ.

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