If int e^{x+tan^{-1} x} left(frac{x^2+2}{1+x^ | vedclass

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

Question

(C) Let $I = \int e^{x+	an^{-1} x} \left(\frac{x^2+2}{1+x^2}ight) dx$.Note that $\sec^2(	an^{-1} x) = 1 + x^2$.Let $g(x) = x + 	an^{-1} x$. Then

Vedclass · Accepted Answer

$f(x)$ is strictly increasing on $\mathbb{R}$.

Vedclass · Answer

(C) Let $I = \int e^{x+	an^{-1} x} \left(\frac{x^2+2}{1+x^2}ight) dx$.Note that $\sec^2(	an^{-1} x) = 1 + x^2$.Let $g(x) = x + 	an^{-1} x$. Then $g'(x) = 1 + \frac{1}{1+x^2} = \frac{1+x^2+1}{1+x^2} = \frac{x^2+2}{1+x^2}$.The integral becomes $\int e^{g(x)} g'(x) dx = e^{g(x)} + c$.Thus, $f(x) = x + 	an^{-1} x$.$f'(x) = 1 + \frac{1}{1+x^2}$.Since $1+x^2 > 0$ for all $x \in \mathbb{R}$, $f'(x) > 0$ for all $x \in \mathbb{R}$.Therefore, $f(x)$ is strictly increasing on $\mathbb{R}$.

If $\int e^{x+\tan^{-1} x} \left(\frac{x^2+2}{1+x^2}\right) dx = e^{f(x)} + c$, then which of the following is true?

Similar Questions

Find the integral of the function $\sin ^{4} x$.

$\int \frac{x - \sin x}{1 - \cos x} dx = $

If $\int \frac{2 \cos x+3 \sin x}{4 \cos x+5 \sin x} dx = \left(\frac{23}{41}\right) x + K \log |4 \cos x+5 \sin x| + c$,then $K$ is equal to

If $\int f(x) \cos x \, dx = \frac{1}{2} [f(x)]^2 + C$ and $f(0) = 0$, then $f'(0) = $

$\int \frac{x}{\sqrt{x^2-2x+5}} dx=$

Vedclass Test Series

Exam Paper Generator

Online Exam Module