Standard +0.8 This question requires executing a non-standard substitution (t² = 2x - 5), algebraic manipulation to verify the given form, then partial fractions and logarithmic integration with definite limits. While methodical, the substitution isn't obvious and requires careful handling of bounds and multiple techniques in sequence, placing it moderately above average difficulty.
9. a. Use the substitution \(t ^ { 2 } = 2 x - 5\) to show that
$$\int \frac { 1 } { x + 3 \sqrt { 2 x - 5 } } \mathrm {~d} x = \int \frac { 2 t } { t ^ { 2 } + 6 t + 5 } \mathrm {~d} t$$
b. Hence find the exact value of
$$\int _ { 3 } ^ { 27 } \frac { 1 } { x + 3 \sqrt { 2 x - 5 } } \mathrm {~d} x$$
Integrates using lns. e.g. \(P\ln(t + 5) + Q\ln(t + 1) \Rightarrow \frac{5}{2}\ln(t + 5) - \frac{1}{2}\ln(t + 1)\)
M1
Uses either the limits 7 and 1 with their attempted integral or alternatively substitutes \(t = (2x - 5)^{\frac{1}{2}}\) and uses the limits 27 and 3 within their attempted integral. Applies the addition law or subtraction law leading to the form \(k\ln a\) or \(\ln b\) where \(a\) and \(b\) are constants
A1
\(\ln 2^{\frac{3}{2}}\) or \(\ln\sqrt{8}\)
(3) marks + (5) marks = (8) marks total for Question 9
**Part a:** Use the substitution $t^2 = 2x - 5$ to show that $\int \frac{1}{x + 3\sqrt{2x - 5}} dx = \int \frac{2t}{t^2 + 6t + 5} dt$
| $t^2 = 2x - 5 \Rightarrow 2t\frac{dt}{dx} = 2 \Rightarrow tdt = dx$ | B1 | $tdt = dx$ or equivalent |
| --- | --- | --- |
| $\int \frac{1}{t^2 + 5 + 3t} dx = \int \frac{1}{t^2 + 5 + 3t} tdt = \int \frac{2t}{t^2 + 5 + 6t} dt$ | M1 | Attempts a full substitution of $t^2 = 2x - 5$ and $x = \frac{t^2 + 5}{2}$, including $dx = tdt$ to form an integrand in terms of $t$ |
| | A1 | Clear reasoning including one fully correct intermediate line, including the integral signs, leading to the given expression |
**Part b:** Hence find the exact value of $\int_3^{27} \frac{1}{x + 3\sqrt{2x - 5}} dx$
| Using partial fractions: $\frac{2t}{t^2 + 6t + 5} = \frac{2t}{(t + 5)(t + 1)} = \frac{A}{t + 5} + \frac{B}{t + 1}$ | | |
| --- | --- | --- |
| $2t = A(t + 1) + B(t + 5)$ | | |
| $t = -5 \Rightarrow -10 = -4A \Rightarrow A = \frac{5}{2}$ | | |
| $t = -1 \Rightarrow -2 = 4B \Rightarrow B = -\frac{1}{2}$ | | |
| $\int \frac{1}{x + 3\sqrt{2x - 5}} = \int \frac{5}{2(t + 5)} - \frac{1}{2(t + 1)} = \frac{5\ln(x + 5)}{2} - \frac{\ln(x + 1)}{2} + c$ | | |
| Finding the new limits: $x = 27 \Rightarrow t = 7, x = 3 \Rightarrow t = 1$ | | |
| $\int_3^{27} \frac{1}{x + 3\sqrt{2x - 5}} dx = \left[\frac{5\ln(7 + 5)}{2} - \frac{\ln(7 + 1)}{2}\right] - \left[\frac{5\ln(1 + 5)}{2} - \frac{\ln(1 + 1)}{2}\right] = \frac{5}{2}\ln(2) + \frac{1}{2}\ln\left(\frac{1}{4}\right)$ | | |
| | $\ln\sqrt{8}$ | |
| b: M1 | Uses correct form of Partial Fraction leading to values of $A$ and $B$ |
| A1 | Correct Partial Fraction $\frac{2t}{t^2 + 6t + 5} = \frac{5}{t + 5} + \frac{-1}{t + 1}$ |
| dM1 | Integrates using lns. e.g. $P\ln(t + 5) + Q\ln(t + 1) \Rightarrow \frac{5}{2}\ln(t + 5) - \frac{1}{2}\ln(t + 1)$ |
| M1 | Uses either the limits 7 and 1 with their attempted integral or alternatively substitutes $t = (2x - 5)^{\frac{1}{2}}$ and uses the limits 27 and 3 within their attempted integral. Applies the addition law or subtraction law leading to the form $k\ln a$ or $\ln b$ where $a$ and $b$ are constants |
| A1 | $\ln 2^{\frac{3}{2}}$ or $\ln\sqrt{8}$ |
**(3) marks + (5) marks = (8) marks total for Question 9**
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