If int frac{e^x}{sqrt{e^{2x}+4e^x+13}} dx = l | vedclass

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

Question

(B) Let $I = \int \frac{e^x}{\sqrt{e^{2x}+4e^x+13}} dx$. Substitute $u = e^x$,then $du = e^x dx$. The integral becomes $I = \int \frac{du}{\sqrt{u^2+4

Vedclass · Accepted Answer

$1$

Vedclass · Answer

(B) Let $I = \int \frac{e^x}{\sqrt{e^{2x}+4e^x+13}} dx$. Substitute $u = e^x$,then $du = e^x dx$. The integral becomes $I = \int \frac{du}{\sqrt{u^2+4u+13}}$. Complete the square for the quadratic expression: $u^2+4u+13 = (u+2)^2 + 9 = (u+2)^2 + 3^2$. Using the standard integral formula $\int \frac{dx}{\sqrt{x^2+a^2}} = \log |x + \sqrt{x^2+a^2}| + c$,we get: $I = \log |(u+2) + \sqrt{(u+2)^2 + 3^2}| + c$. Substituting $u = e^x$ back,we get: $I = \log |e^x + 2 + \sqrt{e^{2x}+4e^x+13}| + c$. Comparing this with the given expression $\log |e^{ax}+2+\sqrt{e^{2x}+4e^x+13}| + c$,we find that $a = 1$.

If $\int \frac{e^x}{\sqrt{e^{2x}+4e^x+13}} dx = \log \left|e^x+2+\sqrt{e^{2x}+4e^x+13}\right|+c$,(where $c$ is the constant of integration),then the value of $a$ in the expression $\log \left|e^{ax}+2+\sqrt{e^{2x}+4e^x+13}\right|+c$ is equal to

Similar Questions

$\int \sin \sqrt{x} \, dx = $

$\int \cos ^3 x \cdot e^{\log (\sin x)} d x=$

The value of the integral $\int \frac{\sin(\ln(2 + 2x))}{x + 1} dx$ is

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

$\int (x+1)(x+3)(x+2)^7 \, dx = $ . . . . . . $+ C$.

Vedclass Test Series

Exam Paper Generator

Online Exam Module