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Question

1(x+1)x2dx

A
23(tan1(x+13))+C
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B
23(tan1(x23))+C
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C
23(tan1(x22))+C
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D
12(tan1(x22))+C
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Solution

The correct option is B 23(tan1(x23))+C
The given form is one of the forms of irrational algebraic functions which is 1(ax+b)cx+ddx.
Now to evaluate such kind of integrals we need to substitute cx+d=t2
So, here we’ll substitute x2=t2
& dx = 2t.dt
So the integral becomes
2t.(t2+3)t.dt
=2(t2+3)dt (use the standard formulae)
=23(tan1(t3))+C
Or 23(tan1(x23))+C (Substituting t = x2)


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