Question

# ∫x+2(x2+3x+3)√x+1dx

A
B
C
D

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=t2−1 & dx = 2t.dt ∫t2−1+2((t2−1)2+3(t2−1)+3)√t2−1+12t.dt ∫(t2+1).2t((t2−1)2+3(t2−1)+3).t 2∫t2+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 -  2∫1+1t2t2+1+1t2dt =2∫1+1t2(t−1t)2+3dt Substitute t−1t=u 1+1t2 .dt = du So, the integral will be = 2∫1(u)2+(√3)2du Which is a standard form. And we can apply the corresponding formula On substituting u = t−1t and using standard formula, the integral becomes 2.1√3tan−1(t−1t√3)+C On substituting t=√x+1 2√3tan−1(x√3(x+1))+CMathematics

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