3x+11−x3=3x+1(1−x)(1+x+x2)3x+1(1−x)(1+x+x2)=A(1−x)+Bx+C(1+x+x2)3x+1=A(1+x+x2)+(Bx+C)(1−x)Put x=1⇒4=3A⇒A=433x+1=A(1+x+x2)+Bx+C−Bx2−Cx3x+1=x2(A−B)+(A+B−C)x+A+CComparing 'x2' coefficientA−B=0⇒A=B⇒B=43 (∵A=43)Comparing 'x' coefficient3=A+B−C⇒3=43+43−C⇒C=8−93⇒C=−13 ∴3x+11−x3=43(1−x)+4x−13(1+x+x2)

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

$\frac{3 x + 1}{1 - x^{3}} =$

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

$\frac{4}{3 (1 + x)} + \frac{4 x - 1}{3 (1 + x - x^{2})}$

$\frac{4}{3 (1 - x)} + \frac{4 x - 1}{3 (1 + x - x^{2})}$

$\frac{4}{3 (1 - x)} + \frac{4 x - 1}{3 (1 + x + x^{2})}$

$\frac{4}{3 (1 + x)} + \frac{4 x - 1}{3 (1 + x + x^{2})}$

answer is C.

(Unlock A.I Detailed Solution for FREE)

Best Courses for You

JEE

NEET

Foundation JEE

Foundation NEET

CBSE

Detailed Solution

$\frac{3 x + 1}{1 - x^{3}} = \frac{3 x + 1}{(1 - x) (1 + x + x^{2})}$

$\begin{array}{l} \frac{3 x + 1}{(1 - x) (1 + x + x^{2})} = \frac{A}{(1 - x)} + \frac{B x + C}{(1 + x + x^{2})} \\ 3 x + 1 = A (1 + x + x^{2}) + (B x + C) (1 - x) \\ Put x = 1 \Rightarrow 4 = 3 A \Rightarrow A = \frac{4}{3} \\ 3 x + 1 = A (1 + x + x^{2}) + B x + C - B x^{2} - C x \\ 3 x + 1 = x^{2} (A - B) + (A + B - C) x + A + C \end{array}$