int frac{x^4+5^{x-1} cdot log _e 5}{x^5+5^x} | vedclass

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(B) Let $I = \int \frac{x^4+5^{x-1} \cdot \log _e 5}{x^5+5^x} \cdot d x$. Consider the substitution $u = x^5 + 5^x$. Then,the derivative is $du = (5x^

Vedclass · Accepted Answer

$\frac{1}{5} \log \left|x^5+5^x\right|$

Vedclass · Answer

(B) Let $I = \int \frac{x^4+5^{x-1} \cdot \log _e 5}{x^5+5^x} \cdot d x$. Consider the substitution $u = x^5 + 5^x$. Then,the derivative is $du = (5x^4 + 5^x \cdot \log_e 5) \cdot dx$. Notice that the numerator of the integrand is $x^4 + 5^{x-1} \cdot \log_e 5$. Since $5^{x-1} = \frac{5^x}{5}$,we can rewrite the numerator as $x^4 + \frac{5^x \cdot \log_e 5}{5} = \frac{5x^4 + 5^x \cdot \log_e 5}{5} = \frac{1}{5} du$. Substituting these into the integral,we get: $I = \int \frac{1}{5} \cdot \frac{du}{u} = \frac{1}{5} \int \frac{du}{u} = \frac{1}{5} \log |u| + C$. Substituting back $u = x^5 + 5^x$,we obtain $I = \frac{1}{5} \log |x^5 + 5^x| + C$. Thus,the correct option is $B$.

$\int \frac{x^4+5^{x-1} \cdot \log _e 5}{x^5+5^x} \cdot d x=$ . . . . . . $+C$.

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