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 $\int \frac{d x}{\cot ^2 x-1}=\frac{1}{A} \log |\sec 2 x+\tan 2 x|-\frac{x}{B}+c$,(where $c$ is the constant of integration),then $A+B=$

$\int \frac{dx}{2+\cos x} = $ (Where $C$ is a constant of integration.)

Let $I(x) = \int \sqrt{\frac{x+7}{x}} \, dx$ and $I(9) = 12 + 7 \log_e 7$. If $I(1) = \alpha + 7 \log_e(1 + 2\sqrt{2})$,then $\alpha^4$ is equal to $..........$.

If $\int \frac{1+x^2}{1+x^4} dx=\frac{1}{\sqrt{2}} \tan ^{-1}\left[\frac{f(x)}{\sqrt{2}}\right]+c$,then $f(x)=$

Let $\int \frac{2-\tan x}{3+\tan x} dx = \frac{1}{2}(\alpha x + \log_e |\beta \sin x + \gamma \cos x|) + C$,where $C$ is the constant of integration. Then $\alpha + \frac{\gamma}{\beta}$ is equal to:

Vedclass Test Series

Exam Paper Generator

Online Exam Module