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

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

Question

(A) Let $I = \int \frac{e^{	an ^{-1} x}}{1+x^2} \left[ (\sec ^{-1} \sqrt{1+x^2})^2 + \cos ^{-1} \left( \frac{1-x^2}{1+x^2} ight) ight] dx$. Since

Vedclass · Accepted Answer

$e^{	an ^{-1} x}(	an ^{-1} x)^2+C$

Vedclass · Answer

(A) Let $I = \int \frac{e^{	an ^{-1} x}}{1+x^2} \left[ (\sec ^{-1} \sqrt{1+x^2})^2 + \cos ^{-1} \left( \frac{1-x^2}{1+x^2} ight) ight] dx$. Since $\sec ^{-1} \sqrt{1+x^2} = 	an ^{-1} x$ and $\cos ^{-1} \left( \frac{1-x^2}{1+x^2} ight) = 2 	an ^{-1} x$,the integral becomes: $I = \int \frac{e^{	an ^{-1} x}}{1+x^2} [(	an ^{-1} x)^2 + 2 	an ^{-1} x] dx$. Let $t = 	an ^{-1} x$,then $dt = \frac{1}{1+x^2} dx$. $I = \int e^t (t^2 + 2t) dt = \int (t^2 e^t + 2t e^t) dt$. Using the formula $\int e^t (f(t) + f'(t)) dt = e^t f(t) + C$,where $f(t) = t^2$ and $f'(t) = 2t$: $I = e^t t^2 + C = e^{	an ^{-1} x} (	an ^{-1} x)^2 + C$.

$\int \frac{e^{\tan ^{-1} x}}{1+x^2}\left[\left(\sec ^{-1} \sqrt{1+x^2}\right)^2+\cos ^{-1}\left(\frac{1-x^2}{1+x^2}\right)\right] d x=$

Similar Questions

$\int_0^1 \frac{x e^x}{(x+1)^2} d x$ is equal to

If $\int e^{\sin x} \cdot \left[ \frac{x \cos^3 x - \sin x}{\cos^2 x} \right] dx = e^{\sin x} f(x) + c$,where $c$ is the constant of integration,then $f(x)$ is equal to:

If $\int \frac{3-x^2}{1-2 x+x^2} e^x d x=e^x f(x)+c$,then find $f(x)$.

If $\int e^x(x^3+x^2-x+4) dx = e^x f(x) + c$,then $f(1) =$

$\int_1^{e} \frac{e^x}{x}(1+x \log x) d x=$

Vedclass Test Series

Exam Paper Generator

Online Exam Module