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The value of
is equal to
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a
4x−3x2+12+38xx2+2+12tan−1x2+C
b
1122x−3x2+22+316x2+2+3162tan−1x2+C
c
2x−3x2+22+38x2+2+122tan−1x2+C
d
none of these
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Correct option is D
x5+x4+4x3+4x2+4x+4=x3x2+2+x2x2+2+2xx2+2+2x2+2=x2+2x3+x2+2x+2=x2+2xx2+2+x2+2=x2+22(x+2). Therefore,f(x)=x5+x4+4x3+4x2+4x+4x2+25=x+2x2+23=122xx2+23+2x2+23Now, ∫dxx2+23=18xx2+22+38∫dxx2+22 using ∫dyy2+k2n∫dxx2+22=14xx2+2+142=tan−1x2+C∴ ∫f(x)dx=−14x2+22+14x2+22+3414xx2+2+142tan−1x2+C=x−14x2+22+316xx2+2+3162tan−1x2+C.Similar Questions
is equal to
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