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Question

Evaluate
i)x2+1x4+1 dx
ii)dxx2+1

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Solution

i)x2+1x4+1dx
=1+1x2x2+1x2dx
=1+1x2(x1x)2+2dx
Let
x1x=t(1+1x2)dx=dt
Therefore, the integral become-
dtt2+(2)2
=12tan1(t2)+C
Substituting the value of t, we get
=12tan1⎜ ⎜ ⎜(1+1x2)2⎟ ⎟ ⎟+C
=12tan1(x2+12x2)+C


ii)1x2+1dx
1x2+(1)2dx
=tan1x+C

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