∫12x2+3x2−4dx is equal to

∫12x2+3x2−4dx=∫1x22+3x2x1−4x2=∫dxx32+3x21−4x2 Let 1−4x2=t24x2=1−t2=∫t4dt2+31−t24t=∫dt8+3−3t2=∫dt11−3t2=∫dt(11)2−(3t)2=121113log⁡11+3t11−3t+C=1233log⁡11+31−4x211−31−4x2+C=1233log⁡11x+3x2−411x−3x2−4+C

∫12x2+3x2−4dx is equal to

∫12x2+3x2−4dx=∫1x22+3x2x1−4x2=∫dxx32+3x21−4x2 Let 1−4x2=t24x2=1−t2=∫t4dt2+31−t24t=∫dt8+3−3t2=∫dt11−3t2=∫dt(11)2−(3t)2=121113log⁡11+3t11−3t+C=1233log⁡11+31−4x211−31−4x2+C=1233log⁡11x+3x2−411x−3x2−4+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

$\int \frac{1}{(2 x^{2} + 3) \sqrt{x^{2} - 4}} d x is equal to$

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

$\frac{1}{2 \sqrt{33}} \log |\frac{\sqrt{11} x - \sqrt{3 (x^{2} + 4)}}{\sqrt{11} x + \sqrt{3 (x^{2} - 4)}}| + C$

$\frac{1}{2 \sqrt{33}} \log |\frac{\sqrt{11} x - \sqrt{3 (x^{2} - 4)}}{\sqrt{11} x + \sqrt{3 (x^{2} - 4)}}| + C$

$\frac{- 1}{2 \sqrt{33}} \log |\frac{\sqrt{11} x + \sqrt{3 (x^{2} + 4)}}{\sqrt{11} x + \sqrt{3 (x^{2} - 4)}}| + C$

$\frac{1}{2 \sqrt{33}} \log |\frac{\sqrt{11} x + \sqrt{3 (x^{2} - 4)}}{\sqrt{11} x - \sqrt{3 (x^{2} - 4)}}| + C$

answer is C.

(Unlock A.I Detailed Solution for FREE)

Best Courses for You

JEE

NEET

Foundation JEE

Foundation NEET

CBSE

Detailed Solution

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

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

$Let 1 - \frac{4}{x^{2}} = t^{2}$

$\begin{array}{l} \frac{4}{x^{2}} = 1 - t^{2} \\ = \int \frac{\frac{t}{4} d t}{[2 + 3 (\frac{1 - t^{2}}{4})] t} \\ = \int \frac{d t}{[8 + 3 - 3 t^{2}]} \\ = \int \frac{d t}{11 - 3 t^{2}} \end{array}$

$\begin{array}{l} = \int \frac{d t}{(\sqrt{11})^{2} - (\sqrt{3} t)^{2}} \\ = \frac{1}{2 \sqrt{11}} \frac{1}{\sqrt{3}} \log |\frac{\sqrt{11} + \sqrt{3} t}{\sqrt{11} - \sqrt{3} t}| + C \\ = \frac{1}{2 \sqrt{33}} \log |\frac{\sqrt{11} + \sqrt{3} \sqrt{1 - \frac{4}{x^{2}}}}{\sqrt{11} - \sqrt{3} \sqrt{1 - \frac{4}{x^{2}}}}| + C \\ = \frac{1}{2 \sqrt{33}} \log |\frac{\sqrt{11} x + \sqrt{3 (x^{2} - 4)}}{\sqrt{11} x - \sqrt{3 (x^{2} - 4)}}| + C \end{array}$