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Question

The value of the integral x3+x+1x21dx is
(where C is an arbitrary constant)

A
x22+ln|x+1|+32ln|x1|+C
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B
x22ln|x+1|+23ln|x1|+C
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C
x22+12ln|x+1|+32ln|x1|+C
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D
x22+ln|x1|+32ln|x+1|+C
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Solution

The correct option is C x22+12ln|x+1|+32ln|x1|+C
It can be seen that the given integrand is not a proper rational expression, therefore, on dividing
x3+x+1x21=x+2x+1x21
Let
2x+1x21=A(x+1)+B(x1)2x+1=A(x1)+B(x+1)(1)
Equating the coefficients of x and constant, we obtain
A+B=2A+B=1
On solving, we obtain
A=12 and B=32

x3+x+1x21=x+12(x+1)+32(x1)x3+x+1x2+1dx=x dx+121(x+1)dx+321(x1)dx=x22+12ln|x+1|+32ln|x1|+C

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