The integral intleft(left(frac{x}{2}right)^x+ | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(A) Let $I = \int \left( \left(\frac{x}{2}\right)^x + \left(\frac{2}{x}\right)^x \right) \log_2 x \, dx$. We know that $\log_2 x = \frac{\ln x}{\ln 2}

Vedclass · Accepted Answer

$\left(\frac{x}{2}\right)^x+\left(\frac{2}{x}\right)^x+C$

Vedclass · Answer

(A) Let $I = \int \left( \left(\frac{x}{2}\right)^x + \left(\frac{2}{x}\right)^x \right) \log_2 x \, dx$. We know that $\log_2 x = \frac{\ln x}{\ln 2}$. Let $f(x) = \left(\frac{x}{2}\right)^x$. Then $\ln f(x) = x \ln \left(\frac{x}{2}\right) = x(\ln x - \ln 2)$. Differentiating both sides with respect to $x$: $\frac{1}{f(x)} f'(x) = 1 \cdot (\ln x - \ln 2) + x \cdot \frac{1}{x} = \ln x - \ln 2 + 1 = \ln x - \ln 2 + \ln e = \ln \left(\frac{ex}{2}\right)$. This does not simplify directly. Let us re-evaluate the derivative of $f(x) = \left(\frac{x}{2}\right)^x$. $f'(x) = \left(\frac{x}{2}\right)^x \left( \ln \left(\frac{x}{2}\right) + 1 \right)$. Now consider $g(x) = \left(\frac{x}{2}\right)^x$. The integral is $\int \left( f(x) + \frac{1}{f(x)} \right) \frac{\ln x}{\ln 2} dx$. Actually,let $u = \left(\frac{x}{2}\right)^x$. Then $du = \left(\frac{x}{2}\right)^x \left( \ln \left(\frac{x}{2}\right) + 1 \right) dx = \left(\frac{x}{2}\right)^x (\ln x - \ln 2 + 1) dx$. This matches the form if we observe that the derivative of $\left(\frac{x}{2}\right)^x$ is $\left(\frac{x}{2}\right)^x (\ln x - \ln 2 + 1)$. By checking the options,the derivative of $\left(\frac{x}{2}\right)^x$ is $\left(\frac{x}{2}\right)^x (\ln x - \ln 2 + 1)$. The integral evaluates to $\left(\frac{x}{2}\right)^x + C$.

The integral $\int\left(\left(\frac{x}{2}\right)^x+\left(\frac{2}{x}\right)^x\right) \log _2 x \, dx$ is equal to

Similar Questions

$\int {x\sqrt {\frac{{1 - {x^2}}}{{1 + {x^2}}}} } \;dx = $

If $\int \frac{d \theta}{\cos ^2 \theta(\tan 2 \theta+\sec 2 \theta)}=\lambda \tan \theta+2 \log |f(\theta)|+C$,where $C$ is a constant of integration,then the ordered pair $(\lambda, f(\theta))$ is equal to

Let $I_{n}(x)=\int_{0}^{x} \frac{1}{(t^{2}+5)^{n}} dt, n=1, 2, 3, \ldots$. Then

Integrate the function: $\frac{5 x+3}{\sqrt{x^{2}+4 x+10}}$

$\int_{\frac{1}{\sqrt[5]{31}}}^{\frac{1}{\sqrt[5]{242}}} \frac{1}{\sqrt[5]{x^{30}+x^{25}}} d x=$

Vedclass Test Series

Exam Paper Generator

Online Exam Module