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Question Number 165561 by LEKOUMA last updated on 03/Feb/22

$$I=\int \frac{dx}{x^8+x^6}$$$$J=\int \frac{1-x^7}{x(1+x^7)}dx$$$$CalculateIandJ$$

Answered by MJS_new last updated on 05/Feb/22

$$\int \frac{dx}{x^8+x^6}=\int (\frac{1}{x^6}-\frac{1}{x^4}+\frac{1}{x^2}-\frac{1}{x^2+1})dx=$$$$=-\frac{1}{5x^5}+\frac{1}{3x^3}-\frac{1}{x}-\arctan x=$$$$=-\frac{15x^4-5x^2+3}{15x^5}-\arctan x+C$$

Answered by MJS_new last updated on 05/Feb/22

$$\int \frac{1-x^7}{x(1+x^7)}dx=-\int \frac{x^7-1}{x(x+1)(x^6-x^5+x^4-x^3+x^2-x+1)}dx=$$$$=\int (\frac{1}{x}-\frac{2}{7(x+1)}-\frac{2(6x^5-5x^4+4x^3-3x^2+2x-1)}{7(x^6-x^5+x^4-x^3+x^2-x+1)})dx=$$$$=\ln x-\frac{2}{7}\ln (x+1)-\frac{2}{7}\ln (x^6-x^5+x^4-x^3+x^2-x+1)=$$$$=\ln \mid x\mid -\frac{2}{7}\ln \mid x^7+1\mid +C$$

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