Mathematics

# $\displaystyle \int x^{x}\left ( 1+\log x \right )dx$ is equal to

$\displaystyle x^{x}+k$

##### SOLUTION
Let $I=\int x^{ x }\left( 1+\log x \right) dx$
Substitute $t={ x }^{ 2 }+1\Rightarrow dt=x^{ x }\left( 1+\log x \right) dx$
$\therefore I=\int { dt } =t+k={ x }^{ x }+k$

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Single Correct Medium Published on 17th 09, 2020
Questions 203525
Subjects 9
Chapters 126
Enrolled Students 84

#### Realted Questions

Q1 Single Correct Medium
Let $f\left ( x \right )$ and $g\left ( x \right )$ be two functions satisfying $f\left ( x^{2} \right )+g\left ( 4-x \right )= 4x^{3},g\left ( 4-x \right )+g\left ( x \right )= 0$, then the value of $\displaystyle \int_{-4}^{4}f\left ( x^{2} \right )dx$ is
• A. $64$
• B. $256$
• C. $0$
• D. $512$

1 Verified Answer | Published on 17th 09, 2020

Q2 Single Correct Medium
$\displaystyle{\int}_{1}^{2}\dfrac{dx}{\left(x^{2}-2x+4\right)^{\tfrac{3}{2}}}=\dfrac{k}{k+5}$, then $k$ is equal to
• A. $2$
• B. $3$
• C. $4$
• D. $1$

1 Verified Answer | Published on 17th 09, 2020

Q3 Single Correct Hard
If $\displaystyle I =\int_{0}^{a} \sqrt {\frac{a-x}{a+x}}dx, a > 0,$ then $I$ equals
• A. $\displaystyle \frac{1}{2}\left(a-\frac{\pi}{2}\right)$
• B. $\displaystyle \frac{a}{2}(\pi-1)$
• C. $\displaystyle \frac{1}{\sqrt{2}}a (\pi-1)$
• D. $\displaystyle a\left( \frac{\pi}{2}-1\right)$

1 Verified Answer | Published on 17th 09, 2020

Q4 Single Correct Hard
Evaluate the integral
$\displaystyle \int_{0}^{\pi/4}\frac{ {s}i {n}\theta+ {c} {o} {s}\theta}{9+16 {s}i {n}2\theta} \ {d}\theta$
• A. $\displaystyle \frac{1}{20} \log 2$
• B. $\log 3$
• C. $\log 2$
• D. $\displaystyle \frac{1}{20} \log 3$

Given that for each $\displaystyle a \in (0, 1), \lim_{h \rightarrow 0^+} \int_h^{1-h} t^{-a} (1 -t)^{a-1}dt$ exists. Let this limit be $g(a)$
In addition, it is given that the function $g(a)$ is differentiable on $(0, 1)$