The correct option is C 12ln(1+tan x3+tan x) + C ∫^2xdx(1+tan x)(3+tan x) Put tan x=t ^2dx· dt ∫dt(1+b)(3+t)·∫A1+t+B3+tdt 3A+B=1 ...

You visited us 1 times! Enjoying our articles? Unlock Full Access!

Property 2

∫2xdx1+tan x3...

Question

$\int \frac{{sec}^{2} x d x}{(1 + tan x) (3 + tan x)}$ .

\frac{1}{2} l n (\frac{2 + tan x}{3 + tan x}) + C

No worries! We‘ve got your back. Try BYJU‘S free classes today!

\frac{1}{2} l n (\frac{3 + tan x}{2 + tan x}) + C

No worries! We‘ve got your back. Try BYJU‘S free classes today!

\frac{1}{2} l n (\frac{1 + tan x}{3 + tan x}) + C

Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses

\frac{1}{2} l n (\frac{1 + tan x}{2 + tan x}) + C

No worries! We‘ve got your back. Try BYJU‘S free classes today!

Open in App

Solution

The correct option is C $\frac{1}{2} l n (\frac{1 + tan x}{3 + tan x}) + C$
$\int \frac{{sec}^{2} x d x}{(1 + tan x) (3 + tan x)}$
Put $tan x = t$
${sec}^{2} d x \cdot d t$
$\int \frac{d t}{(1 + b) (3 + t)} \cdot \int \frac{A}{1 + t} + \frac{B}{3 + t} d t$
$3 A + B = 1$
$A + B = 0$
$= \frac{1}{2} \int \frac{d t}{1 + t} - \frac{1}{2} \int \frac{d t}{3 + t}$
$3 A - A = 1$
$A = \frac{1}{2}, B = - \frac{1}{2}$
$\frac{1}{2} l n (1 + t) - \frac{1}{2} l n (3 + t)$
$= \frac{1}{2} l n (\frac{1 + tan x}{3 + tan x}) + c$ .

Suggest Corrections

Similar questions

\int \frac{dx}{{sin}^{2} x - 12 {cos}^{2} x + cosx sinx}

I : \int \frac{1}{3 + 4 cos x} d x = \frac{1}{\sqrt{7}} l o g [\frac{\sqrt{7} + tan (x / 2)}{\sqrt{7} - tan (x / 2)}] + c

I I : \int \frac{d x}{cos x - sin x} = \frac{1}{\sqrt{2}} log | tan (\frac{x}{2} - \frac{3 π}{8}) | + c

\int (\sqrt{sin x} + \sqrt{cos x})^{- 4} d x =

$\int \frac{1}{2 + 3 s i n x} d x$

\int \frac{(1 + \sqrt{t a n x}) (1 + t a n^{2} x)}{2 t a n x} d x

Join BYJU'S Learning Program