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Answer 1 · 2017-02-23T21:34:01

$int(x^6-1)/(1+x^2)dx=1/5x^5-1/3x^3+x-2arctan(x)+C$

Split up the integral:

$I=int(x^6-1)/(1+x^2)dx=intx^6/(1+x^2)dx-intdx/(1+x^2)$

The second is the arctangent integral:

$I=intx^6/(1+x^2)dx-arctan(x)$

Perform long division on what remains, or write the numerator as follows for the same result:

$I=int(x^4(1+x^2)-x^2(1+x^2)+(x^2+1)-1)/(1+x^2)dx-arctan(x)$

Dividing:

$I=intx^4dx-intx^2dx+intdx-intdx/(1+x^2)-arctan(x)$

$I=1/5x^5-1/3x^3+x-2arctan(x)+C$