If int x^{x}(1+log x) d x=k x^{x}+c,then k= | vedclass

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(A) Let $I = \int x^{x}(1+\log x) dx$. Consider the substitution $u = x^{x}$. Taking the natural logarithm on both sides,$\log u = x \log x$. Differen

Vedclass · Accepted Answer

$\log _{e} e$

Vedclass · Answer

(A) Let $I = \int x^{x}(1+\log x) dx$. Consider the substitution $u = x^{x}$. Taking the natural logarithm on both sides,$\log u = x \log x$. Differentiating with respect to $x$,$\frac{1}{u} \frac{du}{dx} = 1 \cdot \log x + x \cdot \frac{1}{x} = \log x + 1$. Thus,$\frac{du}{dx} = u(1 + \log x) = x^{x}(1 + \log x)$. Therefore,$du = x^{x}(1 + \log x) dx$. Substituting this into the integral,$I = \int du = u + c$. Replacing $u$ with $x^{x}$,we get $I = x^{x} + c$. Comparing this with the given expression $k x^{x} + c$,we find $k = 1$. Since $\log_{e} e = 1$,the correct option is $A$.

If $\int x^{x}(1+\log x) d x=k x^{x}+c$,then $k=$

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