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

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

Question

(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 $B$.

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

Similar Questions

Let $f(x) = \int x^3 \sqrt{3-x^2} dx$. If $5f(\sqrt{2}) = -4$,then $f(1)$ is equal to

$\int \sec^{\frac{2}{3}} x \cdot \operatorname{cosec}^{\frac{4}{3}} x \, dx =$

$\int \frac{\sin x}{\sin (x - \alpha )} dx = $

$\int \frac{1}{x^{\frac{1}{2}}+x^{\frac{1}{3}}} \, dx =$

Let $f(x) = \int \frac{dx}{x^{2/3} + 2x^{1/2}}$ be such that $f(0) = -26 + 24 \log_{e}(2)$. If $f(1) = a + b \log_{e}(3)$, where $a, b \in \mathbb{Z}$, then $a + b$ is equal to:

Vedclass Test Series

Exam Paper Generator

Online Exam Module