Mathematics

# Evaluate $\int x^{2}\log xdx$

##### SOLUTION
$\begin{array}{l} We\, \, have \\ I=\int { { x^{ 2 } }\log x } \\ =\log x.\dfrac { { { x^{ 3 } } } }{ 3 } -\int { \dfrac { { { x^{ 3 } } } }{ 3 } .\dfrac { 1 }{ x } dx } \, \, \, \, \, \left[ { \therefore \int { u.vdx=vu-\int { vdu } } } \right] \\ =\dfrac { { { x^{ 3 } } } }{ 3 } \log x-\dfrac { 1 }{ 3 } \int { { x^{ 2 } }dx } \\ =\dfrac { { { x^{ 3 } } } }{ 3 } \log x-\dfrac { 1 }{ 3 } \times \dfrac { { { x^{ 3 } } } }{ 3 } \\ =\dfrac { { { x^{ 3 } } } }{ 3 } .\log x-\dfrac { { { x^{ 3 } } } }{ 9 } +C \\ It\, is\, the\, required\, answer. \end{array}$

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

#### Realted Questions

Q1 Single Correct Hard
Eavaluate :  $\displaystyle \int \displaystyle x\tan^{-1}x$
• A. $\displaystyle \frac{1}{2}(x^{2}+1)\tan ^{-1}x-\frac{1}{2}x+C$
• B. $\displaystyle \frac{1}{2}x^{2}\tan ^{-1}x-\frac{1}{2}x+ C$
• C. $\displaystyle \frac{1}{2}x^{2}\tan ^{-1}x+\frac{1}{2}x+C$
• D. none of these

1 Verified Answer | Published on 17th 09, 2020

Q2 Single Correct Medium
$\displaystyle\int \left ( \log x \right )^{2}dx.$
• A. $\displaystyle x\left ( \log x \right )^{2}+2x \log x+2x$
• B. $\displaystyle x\left ( \log x \right )^{2}-2x \log x$
• C. $\displaystyle x\left ( \log x \right )^{2}+2x \log x+x$
• D. $\displaystyle x\left ( \log x \right )^{2}-2x \log x+2x$

1 Verified Answer | Published on 17th 09, 2020

Q3 One Word Medium
$I = \dfrac {2}{\pi}\int_{-\pi/4}^{\pi/4} \dfrac {dx}{(1 + e^{\sin x})(2 - \cos 2x)}$ then find $27I^{2}$ equals ________.

1 Verified Answer | Published on 17th 09, 2020

Q4 One Word Medium
If $\displaystyle \int_{0}^{\pi} x f(\cos ^{2}x+\tan ^{4} x) dx=\frac{\pi k}{1997} \int_{0}^{\pi/2} f(\cos^{2}x+\tan^{4} x)dx$ then the value of k is

$\int \frac{1}{1+x}\;dx$