10. (a) Use the substitution $x = u^2 + 1$ to show that \[\int_{1}^{0} \frac{3 dx}{(x - 1)(3 + 2\sqrt{x - 1})} = \int_{u}^{p} \frac{6 du}{u(3 + 2u)}\] 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 Trigonometric Functions: 10. (a) Use the substitution $x

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

Worked Solution & Example Answer:10. (a) Use the substitution $x = u^2 + 1$ to show that \[\int_{1}^{0} \frac{3 dx}{(x - 1)(3 + 2\sqrt{x - 1})} = \int_{u}^{p} \frac{6 du}{u(3 + 2u)}\] 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 $x = u^2 + 1$ to show that

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Answer

To perform the substitution, we start by differentiating:

Let [ dx = 2u , du ]
Substituting in gives us:

[\int \frac{3(2u) du}{(u^2 + 1 - 1)(3 + 2\sqrt{u^2 + 1 - 1})} = \int \frac{6u , du}{u(3 + 2u)}]

By identifying the limits of integration:
When $x=1$ , $u=0$ ; when $x=0$ , $u=-1$ . Therefore, the limits change accordingly.

Now we have: [\int_{0}^{-1} \frac{6u , du}{u(3 + 2u)} = \int_{1}^{0} \frac{6 , du}{3 + 2u}] where $p = 3$ and $q = 2$ .

Step 2

Hence, using algebraic integration, show that

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Answer

Using the result from part (a):
[\int_{1}^{0} \frac{3 dx}{(x - 1)(3 + 2\sqrt{x - 1})} = \ln a]
where we can calculate:

Using partial fractions:
Let [ \frac{6}{u(3 + 2u)} = \frac{A}{u} + \frac{B}{3 + 2u}]

Solving for A and B provides the values for the constants.
After rearranging and integrating, we arrive at:\n[ \text{leading to } k = \ln(\frac{49}{36}) = \ln a]
Thus, we find that $a$ is equal to the rational constant:

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

Use the substitution $x = u^2 + 1$ to show that

Hence, using algebraic integration, show that

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