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Question

x+2(x2+3x+3)x+1dx


A
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B
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C
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D
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Solution

The correct option is C
This is an another form of special irrational algebraic functions. To solve this we’ll substitute x+1=t2 or x=t21
& dx = 2t.dt
t21+2((t21)2+3(t21)+3)t21+12t.dt
(t2+1).2t((t21)2+3(t21)+3).t
2t2+1t4+t2+1.dt
Now we are familiar with the form we have, this is x2±1x4+Kx2+1dx form and to solve this we’ll have to divide numerator and denominator by t2.
So, we'll have - 
21+1t2t2+1+1t2dt
=21+1t2(t1t)2+3dt
Substitute t1t=u
1+1t2 .dt = du
So, the integral will be =
21(u)2+(3)2du
Which is a standard form. And we can apply the corresponding formula

On substituting u = t1t and using standard formula, the integral becomes
2.13tan1(t1t3)+C

On substituting t=x+1
23tan1(x3(x+1))+C

Mathematics

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