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Q.

∫xx9+1x2+1dx=f(x) and f(0)=0 find the value of f(1)

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a

119315−π4+12log2

b

119315−π4+log2

c

283315−π4+log2

d

263315−π4+12log2

answer is D.

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Detailed Solution

∫xx9+1x2+1dx=∫x8−x6+x4−x2+1+x−11+x2dxf(x)=x99−x77+x55−x33+x+12log⁡1+x2−tan−1⁡x+Cf(0)=0⇒C=0∴f(1)=263315−π4+12log⁡2
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∫xx9+1x2+1dx=f(x) and f(0)=0 find the value of f(1)