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

Ans:Hello Student,Please find answer to your question belowUsing partial fraction, we haveOption (3) is correct.

help@askiitians.com

+91-735-322-1155

BACK

Integral Calculus> ∫1-x 7 /x(1+x 7 ) dx equals options ln x ...

∫1-x⁷/x(1+x⁷) dx equals

options

ln x + 2/7 ln (1 + x⁷) + c

ln x – 2/7 ln (1 – x⁷) + c

ln x – 2/7 ln (1 + x⁷) + c

ln x + 2/7 ln (1 – x⁷) + c

Aditya Kartikeya , 11 Years ago

Grade 10

1 Answers

Jitender Singh

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

Last Activity: 11 Years ago