Mathematics

# Solve $\displaystyle\int\dfrac {(1+\log x)^{2}}{x}dx$

##### SOLUTION
$\displaystyle\int\dfrac {(1+\log x)^{2}}{x}dx$
Put, $1+\log x = t$

$\implies \dfrac{1}{x} dx = dt$

Now, $\displaystyle\int \dfrac{(1 + \log x)^2}{x} dx$

$= \int t^{2} dt$

$= \dfrac{t^{3}}{3} + c$

$= \dfrac{(1+\log x)^{3}}{3} + c$

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Subjective Medium Published on 17th 09, 2020
Questions 203525
Subjects 9
Chapters 126
Enrolled Students 86

#### Realted Questions

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solve the following.
$\int\limits_0^1 {\frac{{\log (1 + t)}}{{(1 + {t})^2}}} dt$

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Value of $\displaystyle \int_{-\pi /2}^{\pi /2}\cos ^{3}x \left ( 1+\sin x \right )^{2}\: dx$ is
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Q5 Single Correct Medium
If $y = \frac{1}{x}$, then value of $\int \ ydx$ is
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