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Question

Evaluate: 1x(x2+1)dx.

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Solution

We are given,
I=dxx(x2+1)
Let x=tanθ
dx=sec2θdθ (differentiation)
dx=(1+tan2θ)dθ
dx(1+x2)=dθ
I=dθtanθ
=cotθdθ
I=cosθsinθdθ
Let sinθ=t
cosdθ=dt (differentiation)
I=dtt
I=In|t|+C
I=In|sinθ|+C
Now, sinθ=(tanθ÷secθ)
=x/1+x2
I=In(x1+x2)+C

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