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

A-Level Edexcel Maths Pure Radian Measure: 10. (a) Use the substitution $x

10. (a) Use the substitution $x = u^2 + 1$ to show that $$\int_1^0 \frac{3 dx}{(x-1)(3+2\sqrt{x-1})} = \int_u^3 \frac{6 du}{u(3+2u)}$$ where $p$ and $q$ are positive constants to be found - Edexcel - A-Level Maths Pure - Question 11 - 2020 - Paper 1

Question 11

$10.-(a)-Use-the-substitution-$x-=-u^2-+-1$-to-show-that--$$\int_1^0-\frac{3-dx}{(x-1)(3+2\sqrt{x-1})}-=-\int_u^3-\frac{6-du}{u(3+2u)}$$--where-$p$-and-$q$-are-positive-constants-to-be-found-Edexcel-A-Level Maths Pure-Question 11-2020-Paper 1.png$

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

Worked Solution & Example Answer:10. (a) Use the substitution $x = u^2 + 1$ to show that $$\int_1^0 \frac{3 dx}{(x-1)(3+2\sqrt{x-1})} = \int_u^3 \frac{6 du}{u(3+2u)}$$ where $p$ and $q$ are positive constants to be found - Edexcel - A-Level Maths Pure - Question 11 - 2020 - Paper 1

Step 1

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

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Answer

To solve the integral, we start by using the substitution:

$dx = 2u du$

The limits change accordingly. When $x = 1$ , $u = 0$ ; and when $x = 0$ , $u = -1$ . Therefore, the integral becomes:

$\int_0^{-1} \frac{3(2u) du}{(u^2 + 1 - 1)(3 + 2\sqrt{u^2})}$

This simplifies to:

$\int_0^{-1} \frac{6u du}{u^2(3+2u)}$ .

Step 2

Find $p$ and $q$

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Answer

To find the constants $p$ and $q$ , evaluate the re-written integral:

$\int_0^{-1} \frac{6u du}{u(3+2u)} = \int_0^{-1} \frac{6 du}{3 + 2u}$ .

By completing the integration, we find $p = 3$ and $q = 2$ .

Step 3

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

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Answer

Using the results from part (a), we can conclude:

$\int_1^0 \frac{3 dx}{(x-1)(3 + 2\sqrt{x-1})} = K$

where $K$ integrates to a logarithmic form, specifically:

$K = \ln a \text{, where } a = \frac{49}{36}.$

Thus, we find that the constant $a$ is expressed as a rational number.

10. (a) Use the substitution $x = u^2 + 1$ to show that $$\int_1^0 \frac{3 dx}{(x-1)(3+2\sqrt{x-1})} = \int_u^3 \frac{6 du}{u(3+2u)}$$ where $p$ and $q$ are positive constants to be found - Edexcel - A-Level Maths Pure - Question 11 - 2020 - Paper 1

Worked Solution & Example Answer:10. (a) Use the substitution $x = u^2 + 1$ to show that $$\int_1^0 \frac{3 dx}{(x-1)(3+2\sqrt{x-1})} = \int_u^3 \frac{6 du}{u(3+2u)}$$ where $p$ and $q$ are positive constants to be found - Edexcel - A-Level Maths Pure - Question 11 - 2020 - Paper 1

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

Find $p$ and $q$

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

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