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∫1-x 7 /x(1+x 7 ) dx equals options ln x + 2/7 ln (1 + x 7 ) + c ln x – 2/7 ln (1 – x 7 ) + c ln x – 2/7 ln (1 + x 7 ) + c ln x + 2/7 ln (1 – x 7 ) + c

∫1-x7/x(1+x7)  dx equals
 
options
  1. ln x + 2/7 ln (1 + x7) + c
  2. ln x – 2/7 ln (1 – x7) + c
  3. ln x – 2/7 ln (1 + x7) + c
  4. ln x + 2/7 ln (1 – x7) + c

Grade:10

1 Answers

Jitender Singh IIT Delhi
askIITians Faculty 158 Points
9 years ago
Ans:
Hello Student,
Please find answer to your question below

I = \int \frac{1-x^7}{x(1+x^7)}dx
Using partial fraction, we have
I = \int [\frac{-12x^5+10x^4-8x^3+6x^2-4x+2}{7(x^6-x^5+x^4-x^3+x^2-x+1)}+\frac{1}{x}-\frac{2}{7(x+1)}]dxI = \int [\frac{-12x^5+10x^4-8x^3+6x^2-4x+2}{7(x^6-x^5+x^4-x^3+x^2-x+1)}+log(x)-\frac{2}{7}log(x+1)t = x^6-x^5+x^4-x^3+x^2-x+1
dt = -(-6x^6+5x^4-4x^3+3x^2-2x+1)dx
I = log(x)-\frac{2}{7}log(u)-\frac{2}{7}log(x+1)
I = log(x)-\frac{2}{7}log(x^6-x^5+x^4-x^3+x^2-x+1)-\frac{2}{7}log(x+1)
I = log(x)-\frac{2}{7}log(1+x^7)+c
Option (3) is correct.

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