First slide

Methods of integration

Question

If $\int \frac{d x}{1 + \sqrt[3]{x + 1}} = \frac{3}{2} (x + 1)^{2 / 3} - 3 (x + 1)^{1 / 3}$
$+ f (x) + C then f (x)$ is equal to

Moderate

A

$\log | 1 + \sqrt[3]{x + 1} |$
B

$3 \log | 1 + \sqrt[3]{x + 1} |$
C

$\frac{2}{3} \log | 1 + \sqrt[3]{x + 1} |$
D

$\frac{1}{3} \log | 1 + \sqrt[3]{x + 1} |$

Solution

Put $x + 1 = t^{3}$ , we have
$\int \frac{d x}{1 + \sqrt[3]{x + 1}} = \int \frac{3 t^{2} d t}{1 + t} = 3 \int (t - 1 + \frac{1}{1 + t}) d t$
$\begin{aligned} = 3 [\frac{t^{2}}{2} - t + \log | 1 + t |] + C \\ = \frac{3}{2} (x + 1)^{2 / 3} - 3 (x + 1)^{1 / 3} + \end{aligned}$
$3 \log | 1 + \sqrt[3]{x + 1} | + C$

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