Mathematics

# $\int x log x dx =?$

##### SOLUTION
$\displaystyle \int x\log x d x$
Integrating by parts
$\implies \log x\displaystyle\int x d x-\int \bigg(\frac{d(\log x)}{d x}\bigg)\bigg(\int x d x\bigg)d x=\log x\bigg(\dfrac{x^2}{2}\bigg)-\int \dfrac{1}{x}\times \dfrac{x^2}{2} d x=\dfrac{x^2}{2}\log x-\dfrac{1}{2}\int x d x=\dfrac{x^2}{2}\log x-\dfrac{x^2}{4}+C$

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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
If $\displaystyle I=\int \frac{5x^{8}+7x^{6}}{\left ( x^{2}+1+2x^{7} \right )^{2}}dx$ then $I$ is equal to
• A. $\displaystyle \frac{x^{2}}{2x^{2}+x^{2}+1}+C$
• B. $\displaystyle \frac{x^{5}}{x^{2}+1+2x^{7}}+C$
• C. $\displaystyle \frac{-1}{2x^{7}+x^{2}+1}+C$
• D. $\displaystyle \frac{p\left ( X \right )}{q\left ( x \right )}, deg$ p(x)$=deg$ q(x)$=7$

1 Verified Answer | Published on 17th 09, 2020

Q2 Subjective Hard
Evaluate the given integral.
$\int { \sqrt { 9-{ x }^{ 2 } } } dx$

1 Verified Answer | Published on 17th 09, 2020

Q3 Subjective Hard
Evaluate: $\int_0^\pi {\frac{x}{{1 + \sin \alpha \sin x}}\,\,dx}$

1 Verified Answer | Published on 17th 09, 2020

Q4 Single Correct Medium
The value of $\displaystyle \int_{0}^{\pi }\displaystyle \frac{dx}{1-2\alpha \cos x+\alpha ^{2}}$ is
• A. $\displaystyle \frac{\pi }{1+\alpha ^{2}}$ if $\alpha > 1$
• B. $\displaystyle \frac{\pi }{1+\alpha ^{2}}$ if $\alpha < 1$
• C. $\displaystyle \frac{\pi }{\alpha ^{2}-1}$ if $\alpha < 1$
• D. $\displaystyle \frac{\pi }{\alpha ^{2}-1}$ if $\alpha > 1$

Consider two differentiable functions $f(x), g(x)$ satisfying $\displaystyle 6\int f(x)g(x)dx=x^{6}+3x^{4}+3x^{2}+c$ & $\displaystyle 2 \int \frac {g(x)dx}{f(x)}=x^{2}+c$. where $\displaystyle f(x)>0 \forall x \in R$