$\int \frac{x^{2} (x \sec^{2} x + \tan x)}{(x \tan x + 1)^{2}} d x is equal to$

see full answer

Your Exam Success, Personally Taken Care Of

1:1 expert mentors customize learning to your strength and weaknesses – so you score higher in school , IIT JEE and NEET entrance exams.

An Intiative by Sri Chaitanya

$\frac{x^{2}}{x \tan x + 1} + 2 \log |x \sin x - \cos x| + C$

$\frac{x^{2}}{x \tan x + 1} + 2 \log |x \sin x + \cos x| + K$

$\frac{- x^{2}}{x \tan x + 1} + 2 \log |x \sin x + \cot x| + C$

$\frac{x^{2}}{{(x \tan x + 1)}^{2}} - 2 \log |x \sin x + \cot x| + C$

answer is A.

(Unlock A.I Detailed Solution for FREE)

Best Courses for You

JEE

NEET

Foundation JEE

Foundation NEET

CBSE

Detailed Solution

$Let I = \int \frac{x^{2} (x \sec^{2} x + \tan x)}{(x \tan x + 1)^{2}} d x$

Using integrating by parts

$I = x^{2} \int \frac{(x \sec^{2} x + \tan x)}{{(x \tan x + 1)}^{2}} d x - \int \frac{d}{d x} (x^{2}) \int \frac{(x \sec^{2} x + \tan x)}{{(x \tan x + 1)}^{2}} d x$

$x \tan x + 1 = t$

$\begin{aligned} (x \sec^{2} x + \tan x) d x = d t \\ I = x^{2} \int \frac{d t}{t^{2}} - \int [2 x \int \frac{d t}{t^{2}}] d x \end{aligned}$

$\begin{array}{l} = \frac{- x^{2}}{t} - \int 2 x (\frac{- 1}{t}) d x \\ = \frac{- x^{2}}{x \tan x + 1} + \int \frac{2 x}{(x \tan x + 1)} d x \\ = \frac{- x^{2}}{x \tan x + 1} + \int \frac{2 x \cos x}{x \sin x + \cos x} d x \end{array}$

$Put x \sin x + \cos x = y$

$\begin{array}{l} (x \cos x + 1 \sin x - \sin x) \\ = \frac{- x^{2}}{x \tan x + 1} + 2 \int \frac{d t}{t} \\ = \frac{- x^{2}}{x \tan x + 1} + 2 \log | x \sin x + \cos x | + C \end{array}$