int frac{e^{2025+x} - e^{2025-x}}{e^{2026+x} | vedclass

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

Question

(C) Given integral is $I = \int \frac{e^{2025+x} - e^{2025-x}}{e^{2026+x} + e^{2026-x}} dx$.Factor out $e^{2025}$ from the numerator and $e^{2026}$ fr

Vedclass · Accepted Answer

$\frac{1}{e} \log_e |e^x + e^{-x}|$

Vedclass · Answer

(C) Given integral is $I = \int \frac{e^{2025+x} - e^{2025-x}}{e^{2026+x} + e^{2026-x}} dx$.Factor out $e^{2025}$ from the numerator and $e^{2026}$ from the denominator:$I = \int \frac{e^{2025}(e^x - e^{-x})}{e^{2026}(e^x + e^{-x})} dx = \frac{1}{e} \int \frac{e^x - e^{-x}}{e^x + e^{-x}} dx$.Let $u = e^x + e^{-x}$.Then, the derivative is $du = (e^x - e^{-x}) dx$.Substituting these into the integral:$I = \frac{1}{e} \int \frac{1}{u} du = \frac{1}{e} \ln |u| + C$.Substituting back $u = e^x + e^{-x}$, we get:$I = \frac{1}{e} \ln |e^x + e^{-x}| + C$.

$\int \frac{e^{2025+x} - e^{2025-x}}{e^{2026+x} + e^{2026-x}} dx = $ . . . . . . + $C$

Similar Questions

$\int \sec^{2/3} x \csc^{4/3} x \, dx = $

If $\int \frac{\sin 2x}{\sin^4 x + \cos^4 x} dx = f(x) + c$ where $c$ is the constant of integration, then the value of $\tan(f(x))$ is

Evaluate the integral: $\int \frac{x^4(x^{10}-1)}{x^{20}+3x^{10}+1} dx$

$\int \frac{(1 + \log x)^2}{x} \, dx = $

If $\int {\frac{{\left( {2x + 3} \right)dx}}{{x\left( {x + 1} \right)\left( {x + 2} \right)\left( {x + 3} \right) + 1}}} = C - \frac{1}{{f(x)}}$ where $f(x)$ is of the form of $ax^2 + bx + c$ then $(a + b + c)$ equals

Vedclass Test Series

Exam Paper Generator

Online Exam Module