First slide

Methods of integration

Question

$\int \frac{1}{x {(1 + \log x)}^{3}} d x =$

Easy

A

$- \frac{1}{2 {(1 + \log x)}^{2}} + c$
B

$- \frac{1}{2 {(1 + \log x)}^{3}} + c$
C

$\frac{1}{1 + \log x} + c$
D

$- \frac{1}{3 {(1 + \log x)}^{3}} + c$

Solution

                  $1 + \log x = t$
                  $\frac{1}{x} d x = d t$
        $\int \frac{1}{x {(1 + \log x)}^{3}} d x =$         $\int \frac{1}{t^{3}} d t = \frac{- 1}{2 t^{2}} + c$
                            $= - \frac{1}{2} \frac{1}{{(1 + \log x)}^{2}} + c$

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