∫dx(x+1)2(x2+1) 1(x+1)2 (x2+1)=Ax+1+B(x+1)2+(Cx+D)x2+1 1=A(x+1)(x2+1)+B(x2+1)+(Cx+1)(x+1)2 ⇒x=-1 ⇒ 1=0+B(2)+0 B=12 x=0 ⇒ 1=A+B+D x2coeff. ⇒ 0=A+C x coeff.⇒ 0=A+C+2D D=0 ⇒ A=12 C=-12 I=∫12x+1+12(x+1)2+-12xx2+1dx =12log(x+1)+12(-1)x+1 -14 log(x2+1) =log x+1-12(x+1) -12 logx2+1

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

$\int \frac{d x}{{(x + 1)}^{2} (x^{2} + 1)} =$

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

$\frac{1}{2} \log_{e} \sqrt{x + 1} - \frac{1}{4} \log_{e} \sqrt{x^{2} + 1} + \frac{1}{2 (x + 1)} + c$

$\log_{e} \sqrt{x + 1} - \frac{1}{2} \log_{e} \sqrt{x^{2} + 1} - \frac{1}{2 (x + 1)} + c$

$\frac{1}{4} \log_{e} \sqrt{x + 1} + \frac{1}{2} \log_{e} \sqrt{x^{2} + 1} + \frac{1}{(x + 1)} + c$

$\log_{e} \sqrt{x + 1} + \frac{1}{2} \log_{e} \sqrt{x^{2} + 1} - \frac{1}{x + 1} + c$

answer is B.

(Unlock A.I Detailed Solution for FREE)

Best Courses for You

JEE

NEET

Foundation JEE

Foundation NEET

CBSE

Detailed Solution

$\int \frac{dx}{{(x + 1)}^{2} (x^{2} + 1)} \frac{1}{{(x + 1)}^{2} (x^{2} + 1)} = \frac{A}{x + 1} + \frac{B}{{(x + 1)}^{2}} + \frac{(C x + D)}{x^{2} + 1} 1 = A (x + 1) (x^{2} + 1) + B (x^{2} + 1) + (Cx + 1) {(x + 1)}^{2} \Rightarrow x = - 1 \Rightarrow 1 = 0 + B (2) + 0 B = \frac{1}{2} x = 0 \Rightarrow 1 = A + B + D x^{2} c o e f f . \Rightarrow 0 = A + C x c o e f f . \Rightarrow 0 = A + C + 2 D D = 0 \Rightarrow A = \frac{1}{2} C = \frac{- 1}{2}$

$I = \int (\frac{\frac{1}{2}}{x + 1} + \frac{\frac{1}{2}}{{(x + 1)}^{2}} + \frac{\frac{- 1}{2} x}{x^{2} + 1}) dx = \frac{1}{2} \log (x + 1) + \frac{1}{2} \frac{(- 1)}{x + 1} \frac{- 1}{4} \log (x^{2} + 1) = \log \sqrt{x + 1} - \frac{1}{2 (x + 1)} - \frac{1}{2} \log \sqrt{x^{2} + 1}$