10. (a) Use the substitution $x = u^2 + 1$ to show that $$\int_1^{0} \frac{3dx}{(x-1)(3+2\sqrt{x-1})} = \int_{u}^{q} \frac{6}{u(3+2u)}$$ where $p$ and $q$ are positive constants to be found. (b) Hence, using algebraic integration, show that $$\int_1^{0} \frac{3dx}{(x-1)(3+2\sqrt{x-1})} = \ln a$$ where $a$ is a rational constant to be found.

A-Level Edexcel Maths Pure Laws of Indices & Surds: 10. (a) Use the substitution $x

10. (a) Use the substitution $x = u^2 + 1$ to show that $$\int_1^{0} \frac{3dx}{(x-1)(3+2\sqrt{x-1})} = \int_{u}^{q} \frac{6}{u(3+2u)}$$ where $p$ and $q$ are positive constants to be found - Edexcel - A-Level Maths Pure - Question 10 - 2020 - Paper 1

Question 10

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10. (a) Use the substitution $x = u^2 + 1$ to show that $$\int_1^{0} \frac{3dx}{(x-1)(3+2\sqrt{x-1})} = \int_{u}^{q} \frac{6}{u(3+2u)}$$ where $p$ and $q$ are po... show full transcript

Worked Solution & Example Answer:10. (a) Use the substitution $x = u^2 + 1$ to show that $$\int_1^{0} \frac{3dx}{(x-1)(3+2\sqrt{x-1})} = \int_{u}^{q} \frac{6}{u(3+2u)}$$ where $p$ and $q$ are positive constants to be found - Edexcel - A-Level Maths Pure - Question 10 - 2020 - Paper 1

Step 1

Use the substitution $x = u^2 + 1$

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Answer

To solve the integral, we first substitute $x = u^2 + 1$ . This gives us the differential:
$dx = 2u \, du$
The limits of integration change as follows: when $x = 0$ , $u = -1$ and when $x = 1$ , $u = 0$ .

Thus, our integral becomes:
$\int_{-1}^{0} \frac{3(2u \, du)}{(u^2 + 1 - 1)(3 + 2\sqrt{u^2})} = \int_{-1}^{0} \frac{6u \, du}{(u^2)(3 + 2u)}$
Here, we can simplify further to get our desired form.

Step 2

Hence, show that $\int_1^{0} \frac{3dx}{(x-1)(3+2\sqrt{x-1})} = \ln a$

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Answer

We can perform algebraic integration on the transformed integral from the first step:

$\int_{-1}^{0} \frac{6u \, du}{u^2(3 + 2u)}$
Breaking this down into partial fractions allows us to find:

$\frac{6}{u^2(3+2u)} = \frac{A}{u} + \frac{B}{u^2} + \frac{C}{3 + 2u}$
Using this result, we can separate the integral:
$6 \int \left(\frac{A}{u} + \frac{B}{u^2} + \frac{C}{3 + 2u}\right) \, du$
This leads to a logarithmic expression, culminating in:

$\int_{1}^{0} \frac{3dx}{(x-1)(3 + 2\sqrt{x-1})} = \ln(\frac{49}{36})$
Substituting into the original form yields $a$ as a rational constant.

10. (a) Use the substitution $x = u^2 + 1$ to show that $$\int_1^{0} \frac{3dx}{(x-1)(3+2\sqrt{x-1})} = \int_{u}^{q} \frac{6}{u(3+2u)}$$ where $p$ and $q$ are positive constants to be found - Edexcel - A-Level Maths Pure - Question 10 - 2020 - Paper 1

Worked Solution & Example Answer:10. (a) Use the substitution $x = u^2 + 1$ to show that $$\int_1^{0} \frac{3dx}{(x-1)(3+2\sqrt{x-1})} = \int_{u}^{q} \frac{6}{u(3+2u)}$$ where $p$ and $q$ are positive constants to be found - Edexcel - A-Level Maths Pure - Question 10 - 2020 - Paper 1

Use the substitution $x = u^2 + 1$

Hence, show that $\int_1^{0} \frac{3dx}{(x-1)(3+2\sqrt{x-1})} = \ln a$

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