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Integration of 1/1+x+x²+x³ dx

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Integration of 1/1+x+x²+x³ dx

Posted by Prashant Patel 5 years, 11 months ago

CBSE > Class 12 > Mathematics

1 answers

Sohan .G 5 years, 11 months ago

Split denominator 1+x+x^2+x^3 as =(1+x)+x^2(x+1) =(1+x)+(x^2+1) Now the sum would be like integration of 1/(x+1)(x^2+1)dx No by using partial fraction,formula no.5 (refer pg.317 in ncert textbook) 1/(x+1)(x^2+1)dx=A/(x+1)+Bx+C/(x^2+1).........(1) A/(x+1)+Bx+C/(x^2+1)=A(x+1)+(Bx+C)(x^2+1)/(x+1)(x^2+1) Solving and cancelling the denominator an equating and finding the values of a,b,c gives us A=1/2; B= -1/2; C=1/2 substituting the values of a,b,c in (1) ,splitting the integration terms and integrating gives us the answer ANSWER: (1/2)log|x+1|-(1/4)log|x^2+1|+(1/2)tan^(-1)x+c

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