<ol class="i-list"><li><div class="list-item-content"><span class="list-item-text">Given that $y=\frac{\ln x}{x}$, find $\frac{\mathrm{d}y}{\mathrm{d}x}$ in terms of $x$.</span></div><p style="text-align: right;">[2]</p></li><li><div class="list-item-content"><span class="list-item-text">Hence, or otherwise, find the exact value of $\displaystyle\int\nolimits_{1}^{\mathrm{e}}{\frac{\ln x}{{{x}^{2}}}\mathrm{d}x}$, showing your working.</span></div><p style="text-align: right;">[4]</p></li></ol>

Given that $y=\frac{\ln x}{x}$, find $\frac{\mathrm{d}y}{\mathrm{d}x}$ in terms of $x$.

Hence, or otherwise, find the exact value of $\displaystyle\displaystyle \displaystyle\int\nolimits_{1}^{\mathrm{e}}{\frac{\ln x}{{{x}^{2}}}\mathrm{d}x}$, showing your working.

Techniques of  Integration - General

Techniques of  Integration

International Baccalaureate Higher Level Mathematics

IB HL Math

N2018 P1 Q1

$y = \frac{\ln x}{x}$

$\frac{dy}{dx} = \frac{x \cdot \frac{1}{x} - (\ln x) \cdot 1}{x^2}$

$= \frac{1 - \ln x}{x^2}$

From (i): $\int \frac{1 - \ln x}{x^2} \, dx = \frac{\ln x}{x} + C$

$\int \frac{1}{x^2} - \frac{\ln x}{x^2} \, dx = \frac{\ln x}{x} + C$

$\int x^{-2} \, dx - \int \frac{\ln x}{x^2} \, dx = \frac{\ln x}{x} + C$

$\frac{x^{-1}}{-1} - \int \frac{\ln x}{x^2} \, dx = \frac{\ln x}{x} + C$

$\int \frac{\ln x}{x^2} \, dx = -\frac{1}{x} - \frac{\ln x}{x} - C$

$\int_1^e \frac{\ln x}{x^2} \, dx = \left[-\frac{1}{x} - \frac{\ln x}{x}\right]_1^e$

$= -\frac{1}{e} - \frac{\ln e}{e} - \left[-\frac{1}{1} - \frac{\ln 1}{1}\right]$

$= -\frac{1}{e} - \frac{1}{e} + 1$

$= 1 - \frac{2}{e}$

N2018 P1 Q1

N2018 P1 Q1

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