Let's see who solves it.
Here is a challenge for calculus lovers.
Please try only if you have tried similar problem before.
Don't try to answer for points.
Answers
Answered by
3
it can be solved by Partial fraction method .
1/(1 + x³) = 1/(1 + x)(1 + x² - x)
= A/(1 + x) + (Bx + C)/(1 + x² - x)
now multiply (1 + x)(1 + x² - x) both sides,
1/(1+x³) × (1 + x)(1+x²-x) = A(1+x²-x) + (Bx+C)(1+x)
1 = A + Ax² -Ax + Bx + Bx² + C + Cx
1 = (A+ B)x² + (B + C - A)x² + (A + C)
now compare both sides,
(A + B) = 0
(B + C - A) = 0
( A + C) = 1
solve these equations ,
we get A = 1/3 , B = -1/3 and C = 2/3
so,
1/(1 + x³) = 1/3(1 + x) + (-x + 2)/3(1 + x² - x)
now, see attachment after this ....
the answer will be {ln2}/3 +π/3√3
Thank you
1/(1 + x³) = 1/(1 + x)(1 + x² - x)
= A/(1 + x) + (Bx + C)/(1 + x² - x)
now multiply (1 + x)(1 + x² - x) both sides,
1/(1+x³) × (1 + x)(1+x²-x) = A(1+x²-x) + (Bx+C)(1+x)
1 = A + Ax² -Ax + Bx + Bx² + C + Cx
1 = (A+ B)x² + (B + C - A)x² + (A + C)
now compare both sides,
(A + B) = 0
(B + C - A) = 0
( A + C) = 1
solve these equations ,
we get A = 1/3 , B = -1/3 and C = 2/3
so,
1/(1 + x³) = 1/3(1 + x) + (-x + 2)/3(1 + x² - x)
now, see attachment after this ....
the answer will be {ln2}/3 +π/3√3
Thank you
Attachments:
abhi178:
I hope it's right
Similar questions