Standard +0.3 This is a straightforward substitution problem where the substitution is given explicitly. Students must find du/dx = 2x, rewrite the numerator in terms of u and x, then simplify. While it requires algebraic manipulation to express x³ in terms of u, this is a standard C4 technique with no conceptual surprises, making it slightly easier than average.
or \(6u^{\frac{3}{2}}+16u^{\frac{1}{2}}-4u^{\frac{3}{2}}\) from integration by parts
\(2(x^2-2)^{\frac{3}{2}} + 16(x^2-2)^{\frac{1}{2}} + c\) cao
A1
allow \(2(x^2-2)^{\frac{1}{2}}(x^2+6)+c\); A0 if \(du\) not seen at some stage; must see constant of integration; coefficients must be simplified
## Question 6:
| Answer | Marks | Guidance |
|--------|-------|----------|
| $\frac{du}{dx} = 2x$ or $\frac{dx}{du} = \frac{1}{2}(u\pm 2)^{-\frac{1}{2}}$ | M1 | or substitution of $x = (u\pm 2)^{\frac{1}{2}}$ in denominator from $\frac{dx}{du}$ |
| $\frac{Ax^2 + B}{2}$ or better from replacing $dx$; NB $\frac{6x^3+4x}{2x} = \frac{6x^2+4}{2}$ | M1 | |
| substitution of $x^2 = u\pm 2$ or $x = (u\pm2)^{\frac{1}{2}}$ in numerator | M1 | NB $3(u+2)+2$ or $3(u+2)^{\frac{3}{2}}+2(u+2)^{\frac{1}{2}}$ |
| $\int\left(\frac{3u+8}{\sqrt{u}}\right)[du]$ | A1 | $\frac{3(u+2)+2}{\sqrt{u}}$ or better |
| $\frac{3u^{\frac{3}{2}}}{\frac{3}{2}} + \frac{8u^{\frac{1}{2}}}{\frac{1}{2}}$ | A1 | or $6u^{\frac{3}{2}}+16u^{\frac{1}{2}}-4u^{\frac{3}{2}}$ from integration by parts |
| $2(x^2-2)^{\frac{3}{2}} + 16(x^2-2)^{\frac{1}{2}} + c$ cao | A1 | allow $2(x^2-2)^{\frac{1}{2}}(x^2+6)+c$; A0 if $du$ not seen at some stage; must see constant of integration; coefficients must be simplified |
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