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Question

Let f(x)=x2(1+x2)(1+1+x2)dx and f(0)=0. Then f(1) is equal to

A
loge(1+2)
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B
loge(1+2)π4
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C
loge(1+2)+π4
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D
none of these
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Solution

The correct option is A loge(1+2)π4
Let
f(x)=I=x2(x2+1)(1+x2+1)dx=x2+11x2+1dx=(1x2+11x2+1)dx=I1+I2
Where
I1=1x2+1dx
Substituting x=tantdx=sec2tdt
I1=sectdt=sec2t+tantsecttant+sectdt
Again substituting u=tant+sectdu=(sec2t+tantsect)dt
I1=1u=logu+c=log(x2+1+x)+c
And I2=1x2+1dx=tan1x+c
Hence
f(x)=I=log(x2+1+x)tan1x+Cf(0)=0log1+0+C=0C=0f(1)=log(2+1)π4

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