First slide

Evaluation of definite integrals

Question

The value of the integral $\int_{0}^{3} \frac{d x}{\sqrt{x + 1} + \sqrt{5 x + 1}}$ is

Moderate

A

$11 / 15$
B

$14 / 15$
C

$2 / 5$
D

none of these

Solution

. The integrand can be written as $\frac{\sqrt{x + 1} - \sqrt{5 x + 1}}{- 4 x}$
So the given integral is of the form $I_{1} + I_{2}$
$\begin{array}{l} I_{1} = \frac{- 1}{4} \int_{0}^{3} \frac{\sqrt{x + 1}}{x} d x = - \frac{1}{2} \int_{1}^{2} \frac{t^{2}}{t^{2} - 1} d t \\ = - \frac{1}{2} \int_{1}^{2} (1 + \frac{1}{t^{2} - 1}) d t \\ = - \frac{1}{2} (1 - \frac{1}{2} lim_{t \to 1} \log \frac{| t - 1 |}{t + 1}) \\ (x + 1 = t^{2}) \end{array}$
and $\begin{array}{l} I_{2} = \frac{1}{4} \int_{0}^{3} \frac{\sqrt{5 x + 1}}{x} d x \\ = \frac{1}{4} \int_{1}^{4} \frac{2 u^{2}}{u^{2} - 1} d u (u^{2} = 5 x + 1) \\ = \frac{1}{2} (3 + \frac{1}{2} \log \frac{3}{5} - \frac{1}{2} lim_{u \to 1} \log \frac{| u - 1 |}{u + 1}) \end{array}$
Thus $I_{1} + I_{2} = 1 + \frac{1}{4} \log \frac{3}{5}$

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