int frac{x^2+1}{x^4+7 x^2+1} d x is equal to | vedclass

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

Question

(A) Let $I = \int \frac{x^2+1}{x^4+7 x^2+1} d x$. Divide the numerator and denominator by $x^2$: $I = \int \frac{1+\frac{1}{x^2}}{x^2+7+\frac{1}{x^2}}

Vedclass · Accepted Answer

$\frac{1}{3} 	an ^{-1}\left(\frac{x^2-1}{3 x}ight)+C$

Vedclass · Answer

(A) Let $I = \int \frac{x^2+1}{x^4+7 x^2+1} d x$. Divide the numerator and denominator by $x^2$: $I = \int \frac{1+\frac{1}{x^2}}{x^2+7+\frac{1}{x^2}} d x = \int \frac{1+\frac{1}{x^2}}{(x^2+\frac{1}{x^2}+2)+5} d x$. $I = \int \frac{1+\frac{1}{x^2}}{(x-\frac{1}{x})^2+9} d x$. Let $t = x - \frac{1}{x}$, then $dt = (1 + \frac{1}{x^2}) dx$. Substituting these into the integral: $I = \int \frac{dt}{t^2 + 3^2} = \frac{1}{3} 	an^{-1}(\frac{t}{3}) + C$. Substituting $t = x - \frac{1}{x}$ back: $I = \frac{1}{3} 	an^{-1}(\frac{x - \frac{1}{x}}{3}) + C = \frac{1}{3} 	an^{-1}(\frac{x^2-1}{3x}) + C$.

$\int \frac{x^2+1}{x^4+7 x^2+1} d x$ is equal to

Similar Questions

$\int \frac{\sec x}{3(\sec x+\tan x)+2} d x=$

$\int e^{\tan \theta} (\sec \theta - \sin \theta) d\theta$ equals:

If $\int \sqrt{\frac{x - 5}{x - 7}} dx = A \sqrt{x^2 - 12 x + 35} + \log |x - 6 + \sqrt{x^2 - 12 x + 35}| + C$,then $A = . . . . . .$

The value of $\int {\frac{{{e^x} + 9\cos x - 2\sin x + 7}}{{{e^x} + 7\sin x + 11\cos x + 14}}\,dx} $ is (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

Vedclass Test Series

Exam Paper Generator

Online Exam Module