Mathematics

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

${x^x} + C$

##### SOLUTION

$\int {{x^x}\left( {1 + \log x} \right)} dx$

Let ${x^x} = t$

Applying $log$ on both the side

$log \ x^{x}=log \ t$

$x\log x = \log t$

Taking Derivative w.r.t. $x$

$\left( {x \times \dfrac {1}{x}+ \log x} \right)dx = \dfrac {1}{t}dt$

$\Rightarrow \left( {1 + \log x} \right) \times {x^x}dx = dt$......$\left ( \because t=x^{x} \right )$

$\Rightarrow \int {dt}$

$= t + c$

$= {x^x} + c$

Hence option $C$ is the answer.

Its FREE, you're just one step away

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

#### Realted Questions

Q1 Single Correct Hard
$\displaystyle \int e^{2x} (\sqrt{3}cosx-sinx)dx=$
• A. $\displaystyle \frac{e^{2x}}{5}[(2\sqrt{3}+1)cosx -(\sqrt{3}-2)sinx]+c$
• B. $\displaystyle \frac{e^{2x}}{5}[(2\sqrt{3}+1)sinx+(\sqrt{3}-2)\cos x]+c$
• C. $\displaystyle \frac{e^{2x}}{5}[(2\sqrt{3}+1)sinx-(\sqrt{3}-2)cosx]+c$
• D. $\displaystyle \frac{e^{2x}}{5}[(2\sqrt{3}+1)cosx+(\sqrt{3}-2)sin x]+c$

1 Verified Answer | Published on 17th 09, 2020

Q2 Subjective Medium
Evaluate the integrals
$\int _{ 0 }^{ 2 }{ \cfrac { 1 }{ \left( { x }^{ 2 }+1 \right) } dx }$

1 Verified Answer | Published on 17th 09, 2020

Q3 Single Correct Hard
If $\displaystyle\int { \cfrac { \cos { 8x } +1 }{ \tan { 2x } -\cot { 2x } } } dx=a\cos { 8x } +C\quad$, then $a=$
• A. $-\cfrac { 1 }{ 16 }$
• B. $\cfrac { 1 }{ 8 }$
• C. $-\cfrac { 1 }{ 8 }$
• D. $\cfrac { 1 }{ 32 }$

1 Verified Answer | Published on 17th 09, 2020

Q4 Subjective Hard
Integrate:
$\displaystyle \int{\dfrac{dx}{3x^{2}+13x-10}}$

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 $\displaystyle f(x)>0 \forall x \in R$