If int frac{x^2(x sec^2 x+tan x)}{(x tan x+1) | vedclass

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

Question

(A) Let $I = \int \frac{x^2(x \sec^2 x + 	an x)}{(x 	an x + 1)^2} dx$. Using integration by parts, let $u = x^2$ and $dv = \frac{x \sec^2 x + 	an x

Vedclass · Accepted Answer

$2 \log |x \sin x + \cos x| + c$

Vedclass · Answer

(A) Let $I = \int \frac{x^2(x \sec^2 x + 	an x)}{(x 	an x + 1)^2} dx$. Using integration by parts, let $u = x^2$ and $dv = \frac{x \sec^2 x + 	an x}{(x 	an x + 1)^2} dx$. First, find $v = \int \frac{x \sec^2 x + 	an x}{(x 	an x + 1)^2} dx$. Let $t = x 	an x + 1$, then $dt = (x \sec^2 x + 	an x) dx$. So, $v = \int \frac{1}{t^2} dt = -\frac{1}{t} = -\frac{1}{x 	an x + 1}$. Now, applying the integration by parts formula $\int u dv = uv - \int v du$: $I = x^2 \left( -\frac{1}{x 	an x + 1} ight) - \int \left( -\frac{1}{x 	an x + 1} ight) (2x) dx$. $I = -\frac{x^2}{x 	an x + 1} + \int \frac{2x}{x 	an x + 1} dx$. $I = -\frac{x^2}{x 	an x + 1} + \int \frac{2x \cos x}{x \sin x + \cos x} dx$. Comparing this with the given expression, $f(x) = \int \frac{2x \cos x}{x \sin x + \cos x} dx$. Let $u = x \sin x + \cos x$, then $du = (x \cos x + \sin x - \sin x) dx = x \cos x dx$. Thus, $f(x) = \int \frac{2}{u} du = 2 \log |u| + c = 2 \log |x \sin x + \cos x| + c$.

If $\int \frac{x^2(x \sec^2 x+\tan x)}{(x \tan x+1)^2} dx = \frac{-x^2}{x \tan x+1} + f(x) + c$, then $f(x) =$

Similar Questions

$\int {\frac{{{{\cot }^{ - 1}}({e^x})}}{{{e^x}}}} dx$ is equal to -

$\int \sin ^{-1}\left(\frac{2 x}{1+x^2}\right) d x=$

$I = \int \cos(\ln x) \, dx$. Then $I =$

If $\int {\ln ({x^2} + x)dx = x\ln ({x^2} + x) + A}$,then $A = $

$\int \tan^{-1} \left( \frac{2x}{1 - x^2} \right) dx = $

Vedclass Test Series

Exam Paper Generator

Online Exam Module