Solveeit Logo
Login

Question

Question: Evaluate: \(\int{{{e}^{x}}{{x}^{x}}\left( 2+\log x \right)dx}=\) A. \({{x}^{x}}+c\) B. \({{e}^{...

Evaluate: ∫exxx(2+log⁡x)dx=\int{{{e}^{x}}{{x}^{x}}\left( 2+\log x \right)dx}=
A. xx+c{{x}^{x}}+c
B. exlog⁡x+c{{e}^{x}}\log x+c
C. exxx+c{{e}^{x}}{{x}^{x}}+c
D. ex+xx+c{{e}^{x}}+{{x}^{x}}+c

Explanation

Solution

Hint:We will be using the concepts of integral calculus to solve the problem. We will be using the technique of integration by substitution to solve the problem easily we will first let y=exxxy={{e}^{x}}{{x}^{x}} and differentiate the following with respect to x then we will substitute the value of dx in the given integral.

Complete step-by-step answer:
Now, we have been given an integral,
∫exxx(2+log⁡x)dx\int{{{e}^{x}}{{x}^{x}}\left( 2+\log x \right)dx}
We can rewrite it as,
∫exxx(1+log⁡x+1)dx\int{{{e}^{x}}{{x}^{x}}\left( 1+\log x+1 \right)dx}
Now, we will distribute exxx{{e}^{x}}{{x}^{x}} among (1+log⁡x)\left( 1+\log x \right) and 1. So, we have,
∫(exxx(1+log⁡x)+exxx)dx\int{\left( {{e}^{x}}{{x}^{x}}\left( 1+\log x \right)+{{e}^{x}}{{x}^{x}} \right)dx}
Now, we will substitute exxx{{e}^{x}}{{x}^{x}}.
So, let y=exxxy={{e}^{x}}{{x}^{x}}
Now, on differentiating this we have,
dydx=ddx(exxx)\dfrac{dy}{dx}=\dfrac{d}{dx}\left( {{e}^{x}}{{x}^{x}} \right)
Now, applying product rule of differentiation we have,
=ddx(ex)xx+exddx(xx) dydx=exxx+exddx(xx).............(1) \begin{aligned} & =\dfrac{d}{dx}\left( {{e}^{x}} \right){{x}^{x}}+{{e}^{x}}\dfrac{d}{dx}\left( {{x}^{x}} \right) \\\ & \dfrac{dy}{dx}={{e}^{x}}{{x}^{x}}+{{e}^{x}}\dfrac{d}{dx}\left( {{x}^{x}} \right).............\left( 1 \right) \\\ \end{aligned}
Now, for ddx(xx)\dfrac{d}{dx}\left( {{x}^{x}} \right) we let,
z=xxz={{x}^{x}}
Now, we take logs on both sides. So, we have,
log⁡z=xlog⁡x\log z=x\log x
Now, differentiating both sides we have,
1zdzdx=xx+log⁡x 1zdzdx=1+log⁡x dzdx=z(1+log⁡x) \begin{aligned} & \dfrac{1}{z}\dfrac{dz}{dx}=\dfrac{x}{x}+\log x \\\ & \dfrac{1}{z}\dfrac{dz}{dx}=1+\log x \\\ & \dfrac{dz}{dx}=z\left( 1+\log x \right) \\\ \end{aligned}
Now, re- substituting z=xxz={{x}^{x}} we have,
ddx(xx)=xx(1+log⁡x)\dfrac{d}{dx}\left( {{x}^{x}} \right)={{x}^{x}}\left( 1+\log x \right)
So, from (1) we have,
dydx=exxx+exxx(1+log⁡x) dyexxx+exxx(1+log⁡x)=dx \begin{aligned} & \dfrac{dy}{dx}={{e}^{x}}{{x}^{x}}+{{e}^{x}}{{x}^{x}}\left( 1+\log x \right) \\\ & \dfrac{dy}{{{e}^{x}}{{x}^{x}}+{{e}^{x}}{{x}^{x}}\left( 1+\log x \right)}=dx \\\ \end{aligned}
So, now the integral is on substituting y=xxexy={{x}^{x}}{{e}^{x}} and dydy.

& \int{{{x}^{x}}{{e}^{x}}\left( 2+\log x \right)=}\int{\dfrac{\left( {{e}^{x}}{{x}^{x}}+{{e}^{x}}{{x}^{x}}\left( 1+\log x \right) \right)dy}{\left( {{e}^{x}}{{x}^{x}}+{{e}^{x}}{{x}^{x}}\left( 1+\log x \right) \right)}} \\\ & =\int{dy} \\\ & =y+c \\\ & \int{{{x}^{x}}{{e}^{x}}\left( 2+\log x \right)={{x}^{x}}{{e}^{x}}+c} \\\ \end{aligned}$$ So, the correct option is option c. Note: To solve these types of questions one should know the basic concepts of integral calculus. Also, it is important to note how we have substituted the value of $y={{x}^{x}}{{e}^{x}}$ by realising the fact that the integral given to us is differentiation of this only and hence if we substitute this, the problem will be simplified to a great extent.Students should remember the differentiation formulas i.e $\dfrac{d}{dx}\left( {{e}^{x}} \right)=e^x$ , $\dfrac{d}{dx}\left( \log x\right)=\dfrac{1}{x}$ and product rule of differentiation i.e $\dfrac{d}{dx}\left( uv\right)=u\dfrac{d}{dx}\left( v \right)+v\dfrac{d}{dx}\left( u \right)$ to solve these types of questions.