The value of int e^{tan ^{-1} x} cdot frac{1+ | vedclass

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

Question

(D) Let $I = \int e^{	an ^{-1} x} \cdot \frac{1+x+x^2}{1+x^2} dx$. We can rewrite the integrand as: $I = \int e^{	an ^{-1} x} \left( \frac{1+x^2}{1+

Vedclass · Accepted Answer

$x e^{	an ^{-1} x} + c$

Vedclass · Answer

(D) Let $I = \int e^{	an ^{-1} x} \cdot \frac{1+x+x^2}{1+x^2} dx$. We can rewrite the integrand as: $I = \int e^{	an ^{-1} x} \left( \frac{1+x^2}{1+x^2} + \frac{x}{1+x^2} ight) dx$ $I = \int e^{	an ^{-1} x} \left( 1 + \frac{x}{1+x^2} ight) dx$ $I = \int e^{	an ^{-1} x} dx + \int e^{	an ^{-1} x} \cdot \frac{x}{1+x^2} dx$. Using integration by parts on the first integral $\int e^{	an ^{-1} x} \cdot 1 dx$: Let $u = e^{	an ^{-1} x}$ and $dv = dx$. Then $du = e^{	an ^{-1} x} \cdot \frac{1}{1+x^2} dx$ and $v = x$. Applying the formula $\int u dv = uv - \int v du$: $\int e^{	an ^{-1} x} dx = x e^{	an ^{-1} x} - \int x \cdot e^{	an ^{-1} x} \cdot \frac{1}{1+x^2} dx$. Substituting this back into the expression for $I$: $I = \left( x e^{	an ^{-1} x} - \int \frac{x e^{	an ^{-1} x}}{1+x^2} dx ight) + \int \frac{x e^{	an ^{-1} x}}{1+x^2} dx + c$. $I = x e^{	an ^{-1} x} + c$.

The value of $\int e^{\tan ^{-1} x} \cdot \frac{1+x+x^2}{1+x^2} dx$ is

Similar Questions

If $\int \frac{5 \tan x}{\tan x-2} d x=x+a \log |\sin x-2 \cos x|+c$,then find the value of $a$ (where $c$ is the constant of integration).

If $\int {\frac{{\tan x}}{{1 + \tan x + {{\tan }^2}x}}dx} = x - \frac{K}{{\sqrt A }}{\tan ^{ - 1}}\left( {\frac{{K\tan x + 1}}{{\sqrt A }}} \right) + C,$ ($C$ is a constant of integration),then the ordered pair $(K, A)$ is equal to

$\int \frac{x^{2}+1}{x^{4}+x^{2}+1} d x=$

$\int \frac{25 x^2+8}{\sqrt{25 x^2+9}} d x=$

$\int \frac{3\sin x + 2\cos x}{3\cos x + 2\sin x} \, dx = $

Vedclass Test Series

Exam Paper Generator

Online Exam Module