First slide

Methods of integration

Question

The value of $\int \frac{d x}{\sqrt{x} + \sqrt[3]{x}}$ is

Moderate

A

$3 \sqrt{x} + 3 (\sqrt[3]{x}) - 6 (\sqrt[6]{x}) + 6 \log (\sqrt[6]{x} + 1) + C$
B

$2 \sqrt{x} + 6 \sqrt[6]{x} - 6 \log (\sqrt[6]{x} + 1) + C$
C

$2 \sqrt{x} - 3 (\sqrt[3]{x}) + 6 (\sqrt[6]{x}) - 6 \log (\sqrt[6]{x} + 1) + C$
D

none of these

Solution

We want a substitution that will allow us to
find both the square root and the cube root without getting
fractional exponents. Thus we want a substitution of the
form $x = u^{k}$ where $k$ is a multiple of 2 and 3. Let us use the
least common multiple 6.
Let $x = u^{6}, d x = 6 u^{5} d u$
$\begin{array}{l} \int \frac{d x}{\sqrt{x} + \sqrt[3]{x}} = \int \frac{6 u^{5} d u}{u^{3} + u^{2}} = 6 \int \frac{u^{3}}{u + 1} d u \\ = 6 \int (u^{2} - u + 1 - \frac{1}{u + 1}) d u \\ = 2 u^{3} - 3 u^{2} + 6 u - 6 \log (u + 1) + C \\ = 2 \sqrt{x} - 3 (\sqrt[3]{x}) + 6 (\sqrt[6]{x}) - 6 \log (\sqrt[6]{x} + 1) + C \end{array}$

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