∫x2−1x4+3x2+1tan−1x+1xdx.

I=∫x2−1x4+3x2+1tan−1⁡x+1xdx=∫1−1x2x2+1x2+3tan−1⁡x+1xdx Put x+1x=t∴1−1x2dx=dt and x2+1x2+3=t2+1∴I=∫dtt2+1tan−1⁡t=ln⁡tan−1⁡t+C=lntan−1⁡x+1x+C

∫x2−1x4+3x2+1tan−1x+1xdx.

I=∫x2−1x4+3x2+1tan−1⁡x+1xdx=∫1−1x2x2+1x2+3tan−1⁡x+1xdx Put x+1x=t∴1−1x2dx=dt and x2+1x2+3=t2+1∴I=∫dtt2+1tan−1⁡t=ln⁡tan−1⁡t+C=lntan−1⁡x+1x+C

AI Mentor

Check Your IQ

Free Expert Demo

Try Test

Courses

Dropper NEET Course Dropper JEE Course Class - 12 NEET Course Class - 12 JEE Course Class - 11 NEET Course Class - 11 JEE Course Class - 10 Foundation NEET Course Class - 10 Foundation JEE Course Class - 10 CBSE Course Class - 9 Foundation NEET Course Class - 9 Foundation JEE Course Class -9 CBSE Course Class - 8 CBSE Course Class - 7 CBSE Course Class - 6 CBSE Course

Offline Centres

$\int \frac{x^{2} - 1}{(x^{4} + 3 x^{2} + 1) \tan^{- 1} (x + \frac{1}{x})} d x .$

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

$\log |C o t^{- 1} (x - \frac{1}{x})| + C$

$\log |T a n^{- 1} (x + \frac{1}{x})| + C$

$\log |C o t^{- 1} (x + \frac{1}{x})| + C$

$\log |T a n^{- 1} (x - \frac{1}{x})| + C$

answer is B.

(Unlock A.I Detailed Solution for FREE)

Best Courses for You

JEE

NEET

Foundation JEE

Foundation NEET

CBSE

Detailed Solution

$\begin{array}{l} I = \int \frac{x^{2} - 1}{(x^{4} + 3 x^{2} + 1) \tan^{- 1} (x + \frac{1}{x})} d x \\ = \int \frac{1 - \frac{1}{x^{2}}}{(x^{2} + \frac{1}{x^{2}} + 3) \tan^{- 1} (x + \frac{1}{x})} d x \\ Put x + \frac{1}{x} = t \\ ∴ (1 - \frac{1}{x^{2}}) d x = d t and x^{2} + \frac{1}{x^{2}} + 3 = t^{2} + 1 \\ ∴ I = \int \frac{d t}{(t^{2} + 1) \tan^{- 1} t} = \ln |\tan^{- 1} t| + C \\ = l n |\tan^{- 1} (x + \frac{1}{x})| + C \end{array}$