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

$\int {\frac{{{x^2} - 1}}{{{x^4} + {x^2} + 1}}\,dx = }$

$\int_{\frac{1}{\sqrt[5]{31}}}^{\frac{1}{\sqrt[5]{242}}} \frac{1}{\sqrt[5]{x^{30}+x^{25}}} d x=$

$\int \frac{dx}{4+3 \cot x} = $

For $-1 < x, y < 1$, if $\int \frac{x}{\sqrt{1+x}+\sqrt{1-x}} dx + \int \frac{y}{\sqrt{y+1}+\sqrt{y-1}} dy = A(1+x)^{3/2} + B(1-x)^{3/2} + f(y)(y+1)^{3/2} + g(y)(y-1)^{3/2} + C$, then $A f(y) + B g(y) =$

If $\int \frac{dx}{(x-1)^{3/2}(x-3)^{1/2}} = \sqrt{f(x)} + c$,then $f(-1) - f(0) =$

Vedclass Test Series

Exam Paper Generator

Online Exam Module